Permutations and

01 Combinations

Introduction
Counting is the most fundamental application of mathematics. There are many natural methods used for
counting.
In this chapter we will be going to deal with various known techniques those are much faster than the usual
counting methods.

Important Point
𝐹𝑃𝐶 (Fundamental Principal of Counting) is used to count some event without actually counting them.
Let us take help of some model.
Model- I :
Find number of ways of in which one can travel from 𝑇1 (town1) to 𝑇3 (town3) via 𝑇2 (town2).
𝑅1 𝑅4
𝑅2
𝑇1 𝑇2 𝑇3
𝑅3
𝑅5

Total ways :-

𝑇1 𝑅1 𝑇2 𝑅4 𝑇3

𝑇1 𝑅1 𝑇2 𝑅5 𝑇3

𝑇1 𝑅2 𝑇2 𝑅4 𝑇3

𝑇1 𝑅2 𝑇2 𝑅5 𝑇3

𝑇1 𝑅3 𝑇2 𝑅4 𝑇3

𝑇1 𝑅3 𝑇2 𝑅5 𝑇3 = 6 ways

It is easy to proceed by 𝐹𝑃𝐶 𝑇1 to 𝑇2 ⎯→ 3

𝑇2 to 𝑇3 ⎯→ 2

Total ways = 3 × 2 = 6

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Model- II:
To find the number of ways by which a person can enter and leave cinema hall by a different door.
4 + 4 + 4 + 4 + 4 = 20
𝐷2
𝐷3
𝐷1
𝐷4
𝐷5
Cinema Hall 𝐷4
𝐷2 𝐷3
𝐷1 𝐷4
𝐷5 𝐷5
𝐷1
Cinema
𝐷3 𝐷2
Hall 𝐷4
𝐷2 𝐷4 𝐷5
𝐷1
𝐷3 𝐷2
𝐷4
𝐷3
𝐷5
𝐷1
𝐷5 𝐷2
𝐷3
𝐷4
By 𝐹. 𝑃. 𝐶.
(i) A person can enter in cinema hall by 5 ways & leave by 4 ways = 5 × 4 = 20.
(ii) If he can enter and leave by any door then number of ways = 5 × 5 = 25.
(iii) He can enter by 𝐷1 , 𝐷2 and leaves by 𝐷3 , 𝐷4 , 𝐷5 = 2 × 3 = 6
(iv) He enters with odd number gate and leaves the even number gate = 3 × 2 = 6
Basic Steps to Remember :
Step-I : Identify the independent events involved in a given problem.
Step-II : Find the number of ways performing/occurring each event
Step-III : Multiply these numbers to get the total number of ways of performing/occurring all the events
Example :
There are 15 IITs in India and let each IIT has 10 branches, then the IITJEE topper can select the IIT and
branch in 15 × 10 = 150 number of ways.
Example :
There are 15 IITs & 20 NITs in India, then a student who cleared both IITJEE & AIEEE exams can select an
institute in (15 + 20) = 35 number of ways.
Illustration 1:
Number of ways in which IITJEE topper can select the IIT and its branch, if there are 23 IITs in India and
each IIT has 10 branches.
Solution:
Number of ways = 23 × 10 = 230
Illustration 2:
Number of ways in which a student who cleared both IITJEE & AIEEE exams can select an institute if there
are 23 IITs & 31 NITs in India is :
Solution:
Number of ways = 23 + 31 = 54

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Permutations and Combinations
Fundamental Counting Problems
Illustration 3:
Shubham has 2 school bags, 3 tiffin boxes and 2 water bottles. In how many ways can she carry these items
(choosing one each).
Solution:
A school bag can be chosen in 2 different ways. After a school bag is chosen, a tiffin box can be chosen in 3
different ways. Hence, there are 2 × 3 = 6 pairs of school bag and a tiffin box. For each of these pairs a
water bottle can be chosen in 2 different ways.
Hence, there are 6 × 2 = 12 different ways in which, Shubham can carry these items to school. If we name
the 2 school bags as 𝐵1 , 𝐵2 , the three tiffin boxes as 𝑇1 , 𝑇2 , 𝑇3 and the two water bottles as 𝑊1 , 𝑊2 these
possibilities can be illustrated in the Figure.
12 Possibilities
𝑊1 𝐵1 𝑇1 𝑊1
𝑇1 𝑊2
𝐵1 𝑇1 𝑊2
𝑊1 𝐵1 𝑇2 𝑊1
𝑇2
𝑊2
𝑇3 𝐵1 𝑇2 𝑊2
𝐵1 𝑊1 𝐵1 𝑇3 𝑊1
𝑊2
𝐵1 𝑇3 𝑊2
𝑊1 𝐵2 𝑇1 𝑊1
𝑊2
𝐵2 𝑇1 𝐵2 𝑇1 𝑊2
𝑊1 𝐵2 𝑇2 𝑊1
𝑇2
𝑊2
𝑇3 𝐵2 𝑇2 𝑊2
𝑊1 𝐵2 𝑇3 𝑊1
𝑊2
𝐵2 𝑇3 𝑊2
Illustration 4:
A college offers 6 courses in the morning and 4 in the evening. The possible number of choices with the
student if he wants to study one course in the morning and one in the evening is-
(A) 24 (B) 2 (C) 12 (D) 10
Ans. (A)
Solution:
The student has 6 choices from the morning courses out of which he can select one course in 6 ways.
For the evening course, he has 4 choices out of which he can select one in 4 ways.
Hence the total number of ways 6 × 4 = 24.
Illustration 5:
A college offers 6 courses in the morning and 4 in the evening. The number of ways a student can select
exactly one course, either in the morning or in the evening-
(A) 6 (B) 4 (C) 10 (D) 24
Ans. (C)
Solution:
The student has 6 choices from the morning courses out of which he can select one course in 6 ways.
For the evening course, he has 4 choices out of which he can select one in 4 ways.
Hence the total number of ways 6 + 4 = 10.

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Illustration 6:
Tossing of a coin & Tree diagram
Solution:
𝑇
𝑇
𝐻
𝑇
𝑇
𝐻 8 ways
𝐻
Start (2 × 2 × 2)
𝑇
𝑇
𝐻
𝐻
𝑇
𝐻
𝐻
Illustration 7:
In an examination of 10 𝑇/𝐹 question, How many sequence of answers are possible.
Solution:
Any question can be answered in two ways , i.e. true or false.
So total task of answering tan question can be done in
2 × 2 × 2 × . . . . . . . . . . . 10 times = 210 ways
Illustration 8:
10 students complete in a swimming race. In how many ways can they occupy the first 3 positions.
Solution:
1st place can be occupied in 10 ways
2nd place can be occupied in 9 ways
3rd place can be occupied in 8 ways.
So total number of ways = 10 × 9 × 8 = 720
Illustration 9:
There are 7 flags of different colour. Find the number of different signals that can be transmitted by the
use of 2 flags one above the other.
Solution:
1st place can be occupied in 7 ways
2nd place can be occupied in 6 ways
So total number of ways = 7 ∙ 6 = 42

Fundamental Principle of Counting

Multiplication Principle (Fundamental Principle of Counting)
Suppose an event 𝐸 can occur in 𝑚 different ways and associated with each way of occurring of 𝐸, another
event 𝐹 can occur in 𝑛 different ways, then the total number of occurrences of the two events in the given
order is 𝑚 × 𝑛.
Addition Principle
If an event 𝐸 can occur in 𝑚 ways and another event 𝐹 can occur in 𝑛 ways, and suppose that both cannot
occur together, then 𝐸 or 𝐹 can occur in 𝑚 + 𝑛 ways.

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Permutations and Combinations
Notation of Factorial
1. Notation of Factorial & its Algebra
The continued product of first 𝑛, natural number is called as "𝑛 factorial" and denoted by 𝑛!
𝑛! = 𝑛. (𝑛 – 1). (𝑛 – 2). . . . . . . 3.2.1
4! = 4.3.2.1 = 24
3! = 3.2.1 = 6
5! = 120 ; 6 ! = 720 ; 7 ! = 5040
Special Results :
(I) 0! = 1 i.e. factorial of zero is 1
Proof : 𝑛! = 𝑛. (𝑛 – 1)!
Putting 𝑛 = 1
1! = 1.0! ⇒ 0! = 1
(II) Factorial of negative number is undefined
n! 0! 1
( n − 1)! = if 𝑛 = 0 then (−1)! = = Not defined
n 0 0
Illustration 10:
Find 𝑛 if (𝑛 + 1)! = 12 × (𝑛 – 1)!
Solution:
(𝑛 + 1) 𝑛(𝑛 – 1)! = 12 × (𝑛 – 1)!
𝑛2 + 𝑛 – 1 2 = 0; (𝑛 + 4) (𝑛 – 3) = 0 ∴ 𝑛 = 3
Illustration 11:
(𝑛 + 2)! = 2550 𝑛!
Solution:
(𝑛 + 2) (𝑛 + 1) = 2550; (𝑛 + 52) (𝑛 – 4 9) = 0 ∴ 𝑛 = 49

2. Exponent of Prime Number (P) in n!

Let 𝑝 be a prime number and 𝑛 be a positive integer. Then, the last integer amongst 1, 2, 3,........
𝑛 𝑛 𝑛
(𝑛 – 1), 𝑛 which is divisible by 𝑝 is [ ] 𝑝, where [ ] denotes the greatest integer less than or equal to
𝑝 𝑝 𝑝
 10   12   15 
For example,   = 3,   = 2,   = 5 etc.
3 3 3
n  n   n  n 
E p ( n !) =   +  2  +  3  ++  s 
 p  p   p  p 
Where 𝑠 is the largest positive integer such that p s  n  p s +1
 100   100   100   100   100   100 
E2 (100!) = E2 (100!) =  + 2 + 3 + 4 + 5 + 6 
 2   2   2   2   2   2 
= 50 + 25 + 12 + 6 + 3 + 2
Illustration 12:
Exponent of 3 in 100! Is equal to.
Solution:
 100   100   100   100 
E3 = E3 =  + + + 
 3   9   27   81 
= [33.3] + [11.1] + [3.7] + [1.2] = 33 + 11 + 3 + 1 = 48

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Illustration 13:
Find number of zeros at the end of (1000)!
Solution:
In any usual factorial of a natural number of 2𝑠 are more than number of 5𝑠. Hence number of 10𝑠 are same
as number of 5𝑠.
Objective approach:
 1000   1000   1000   1000 
E5 (1000!) =  + 2 + 3 + 4 
 5   5   5   5 
= 200 + 40 + 8 + 1 = 249

Forming Numbers and Arranging Digits

Divisibility of Numbers:
The following chart shows the conditions of divisibility of numbers.
Divisible by Condition
2Whose last digit is even (0, 2, 4, 6, 8)
4Whose last two digits number is divisible by 4
8Whose last three digits number is divisible by 8
3Sum of whose digits is divisible by 3
9Sum of whose digits is divisible by 9
6Which is divisible by both 2 and 3
5Whose last digit is either 0 or 5
25Whose last two digits are divisible by 25
11 Difference of sum of digits at odd place with even place should be divisible by 11.
10Divisible by 2 and 5
Illustration 14:
How many 3 digits numbers can be formed by the digit 1, 2, 3, 4, 5 without repetition.
Solution:
Hundred's place digit can be selected in 5 ways.
Ten's place digit can be selected in 4 ways.
Unit's place digit can be selected in 3 ways.
So, 5 × 4 × 3 = 60
Illustration 15:
How many four digits numbers can be made by using 0, 1, 3, 4, 7, 9
(i) If repetition allowed
(ii) If repetition not allowed
(iii) Even Numbers (Repetition allowed)
(iv) Odd (Repetition allowed)
Solution:
Given digits: 0, 1, 3, 4, 7, 9 → 6 digits
(i)

   
5 × 6 × 6 × 6 = 1080
(zero not include)

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(ii)

   
5 × 5 × 4 × 3 = 300

(zero not include)

(iii)
0 or 4
   
5 × 6 × 6 × 2 = 360

(zero not include)

(iv)
1 or 3 or 7 or 9
   
5 × 6 × 6 × 4 = 720

Illustration 16:
How many 6 digits odd number greater than 6,00,000 can be formed from the digits 5,6,7, 8, 9, 0 if
repetition of digit is allowed?
Solution:
Numbers greater than 6,00,000 and formed with the digit 5, 6, 7, 8, 9, 0 are of 6 digit but begin with 6, 7, 8 or 9.
Also, the numbers which end with 5, 7, 9 are odd.
Hence, first place can be filled by 4 ways (out of 6, 7, 8 or 9). Last place can be filled by 3 ways.
Hence, first and last place can be filled by 4 × 3 ways.
Also 2nd place can be filled by 6 ways.
3rd place can be filled by 6 ways
4th place can be filled by 6 ways.
5th place can be filled by 6 ways
Hence, all the 6 places can be filled by
4 × 3 × 6 × 6 × 6 × 6 = 15552 ways.

Definition of Permutation & Combination:

Permutation:
Permutation means arrangement in a definite order of things which may be alike or different taken some
or all at a time. Hence permutation refers to the situation where order of occurrence of the events is
important.
Example:
(i) Out of 𝐴, 𝐵, 𝐶, 𝐷 take 3 letters & form number plate of car.
(ii) Selection of cricket team of 11 players from 16 players is combination but deciding with batting order
is permutation.
Theorm-1 :
Number of permutations of 𝑛 distinct things taken all at a time symbolised as :
Pn = P ( n , n ) = Ann = n !
n

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Proof :
Let these are 𝑛 things arranged at 𝑛 places
n. ( n –1) . ( n –2) .........3.2.1 = n!
We also say that number of ways in which n distinct objects can be arranged amongst themselves in
Pn = n ! i.e. Find total number of words that of 10 letters that can be formed from all the letters of word
n

GANESHPURI.
n
A = Pn = 10! = n !
Theorm-2 :
Number of permutations of 𝑛 distinct things taken 𝑟 at a time
0  r  n
n n!
Pr = P( n , r ) = Arn =
( n − r )!
Things 𝑇1 , 𝑇2 . . . . . . 𝑇𝑛
Places 1, 2, 3 . . . . . . . . . . . 𝑟
Choice 𝑛. (𝑛 − 1). (𝑛 − 2) . . . . . . . . [𝑛 − (𝑟 − 1)]
n( n − 1)( n − 2)( n − r + 1)( n − r )! n!
Total way = =
( n − r )! ( n − r )!
Hence, we can say that
n n!
Pr =
( n − r )!

= ( n )( n –1)( n –2) ....... ( n – r + 1)


r factors
Note :
1. 100
P = 100  99
2
n
2. P1 = n
n n
3. P0 = =1
n
Illustration 17:
Simplify:
3 10 100
i. P ii. P iii. P
2 5 2

Solution:
3!
P= =6
3
i.
2
( 3 − 2) !
10! 10.9.8.7.6.5!
P = = = 10.9.8.7.6
10
ii.
5
(10 − 5)! 5!

iii. 100
P2 = 100  99

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Permutations and Combinations
Illustration 18:
In how many ways can 5 persons be made to occupy five different chairs.
Solution:
P5 = 5! = 120
5

Illustration 19:
In how many ways can 5 persons be made to occupy three different chairs.
Solution:
P3 = 5  4  3 = 60
5

Illustration 20:
In how many ways first three prizes are distributed in seven athletes?
Solution:
M-1 : P1 P2 P3

7 × 6 × 5  210

7
M-2 : P3
= 7. 6. 5 = 210
Combination:
Combination/selection/collection/committee refers to the situation where order of occurrence of the
event is not important. Combination is selection of one or more things out of 𝑛 things which may be alike
or different.
Note:
Things which are alike and which are different. All god made things in general are treated to be different
and all man made things are to be spelled whether like or different.
Hence, we say that permutation is arrangement of things in definite order.
Example:
(i) Out of four letters 𝐴, 𝐵, 𝐶, 𝐷 take any 3 letters & form triangle (possible).
Theorm-1 :
Number of combination/selections of 𝑛 distinct things taken 𝑟 at a time
n n!
n
Cr = c ( n , r ) =   =
 r  ( r )! ( n − r )!
Proof:
Let 10 different objects are given as 𝐴, 𝐵, 𝐶, 𝐷, 𝐸, 𝐹, 𝐺, 𝐻, 𝐼, 𝐽
Let combinations taking 3 at a time = 𝑥
Arrangement = ( x )  (3!)
10
x.3! = P3
10
P3 10!
x= =
3! (10 − 3)! 3!

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Illustration 21:
Find the value of
10
i. 5 C2 ii. 10
C2 iii. C3
Solution:
5 5 5.4
i. C2 = = = 10
5 − 2 2 2
10 10 10  9  8 10  9
ii. C2 = = = = 45
8 2 8 2 2
10 10  9  8
iii. C3 =
3
Keep in Mind :
n
❖ C0 = 1
n
❖ C1 = n
n
❖ Cn = 1
Illustration 22:
Find the value of 𝑛 such that 𝑛 𝑃5 = 42 𝑛 𝑃3 , 𝑛 > 4
Solution:
Given that
𝑛
𝑃5 = 42 𝑛 𝑃3
or 𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)(𝑛 − 4) = 42𝑛(𝑛 − 1)(𝑛 − 2)
Since 𝑛 > 4 so 𝑛(𝑛 − 1)(𝑛 − 2) ≠ 0
Therefore, by dividing both sides by 𝑛(𝑛 − 1)(𝑛 − 2), we get
(𝑛 − 3(𝑛 − 4) = 42
or 𝑛2 − 7𝑛 − 30 = 0
or 𝑛2 − 10𝑛 + 3𝑛 − 30 = 0
or (𝑛 − 10)(𝑛 + 3) = 0
or 𝑛 − 10 = 0 or 𝑛 + 3 = 0
or 𝑛 = 10 or 𝑛 = −3
As 𝑛 cannot be negative, so 𝑛 = 10.
Illustration 23:
n
P4 5
Find the value of 𝑛 such that n −1
= ,n  4
P4 3
Solution:
n
P4 5
Given that n −1
=
P4 3
Therefore 3𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3) = 5(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)(𝑛 − 4)
or 3𝑛 = 5(𝑛 − 4)[𝑎𝑠(𝑛 − 1)(𝑛 − 2)(𝑛 − 3) ≠ 0, 𝑛 > 4] or 𝑛 = 10.
Illustration 24:
Find 𝑟, if 5 4 𝑃𝑟 = 6 5 𝑃𝑟−1 .

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Solution:
We have 5 4 𝑃𝑟 = 6 5 𝑃𝑟−1
4! 5!
Or 5  = 6
( 4 − r ) ! ( r + 1)!
5 −
5! 6  5!
Or =
( 4 − r ) (
! 5 − r + 1 )( − r )(5 − r − 1)!
5
or (6 − 𝑟)(5 − 𝑟) = 6
or 𝑟 2 − 11𝑟 + 24 = 0
or 𝑟 2 − 8𝑟 − 3𝑟 + 24 = 0
or (𝑟 − 8)(𝑟 − 3) = 0
or 𝑟 = 8 or 𝑟 = 3.
Hence 𝑟 = 8, 3.

Rank Problem or Dictionary Problems

Lexicography : The theory and practising of writing and editing dictionary is known as LEXICOGRAPHY
Note :
RANK of the word means, the no. from the starting at which the required word occurs in a special
dictionary which is formed by using all the letters of the given word.
Illustration 25:
Find rank of the word ′𝐶𝐴𝐵′.
Solution:
A, B, C
𝐴 = 2

𝐵 = 2

𝐶 𝐴 𝐵 = 1

Rank = 5
Illustration 26:
Find total number of 5 letter word that can be formed from letters of word "TOUGH".
Solution:

5 × 4 × 3 × 2 × 1 = 120
Illustration 27:
Find the rank of "TOUGH" if all the letters of the word are arranged in all possible orders & written out as
in a dictionary.
Solution:
The number of letters in the word "TOUGH" is 5 & all the five letters are different.
Alphabetical order of all the letters is G,H,O,T,U
Number of words beginning with G =4×3×2×1
Number of words beginning with H =4×3×2×1
Number of words beginning with O =4×3×2×1
Number of words beginning with TG = 3 × 2 × 1
Number of words beginning with TH = 3 × 2 × 1
Number of words beginning with TOG = 2 × 1
Number of words beginning with TOH = 2 × 1
Next words beginning with "TOU" and it is "TOUGH" = 1.
Rank = 24 + 24 + 24 + 6 + 6 + 2 + 2 + 1 = 89

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Illustration 28:
Find rank of the word 'BIHAR'.
Solution:
A, B, H, I, R
𝐴 = 4 = 24

𝐵 𝐴 = 3=6

𝐵 𝐻 = 3=6

𝐵 𝐼 𝐴 = 2=2

𝐵 𝐼 𝐻 𝐴 𝑅 = 1

Rank = 24 + 6 + 6 + 2 + 1 = 39

Problems based on words

Illustration 29:
Number of words which can be made by using all the letters of the word 𝐽𝑂𝐷𝐻𝑃𝑈𝑅.
(i) If no condition.
(ii) If word starts with ′𝐽′ and ends with ′𝑅′.
(iii) If word starts with ′𝐽′ or ends with ′𝑅′.
(iv) If word neither starts with ′𝐽′ nor ends with ′𝑅′.
(v) If all vowels occupy odd places.
(vi) If position of vowels or consonants remain same.
Solution:
(i) 𝐽, 𝑂, 𝐷, 𝐻, 𝑃, 𝑈, 𝑅 → 7
7 × 6 × 5 × 4 × 3 × 2 × 1 = 17
Number of words are 7! = 5040
(ii) Starts with ′𝐽′ and ends with ′𝑅′ then the number are comes in middle part is 5!.

      
1 × 5 × 4 × 3 2 × 1 × 1 = 120

(iii) (Starts with ′𝐽′) + (End with ′𝑅′)(Start 𝐽 and End 𝑅)

OR
𝐽 𝑅 𝐽 𝑅
( ……….) + (…………. )–( ………… )

𝐴 or 𝐵 = 𝐴 ∪ 𝐵
OR
= 𝐴 + 𝐵 − (𝐴 ∩ 𝐵)

(1 × 6!) + (6! × 1)– (5!)

6! + 6!– 5! = 1320

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Permutations and Combinations
(iv) Total words – [Case-iii]
= 5040– 1320 = 3720
(v)
1 2 3 4 5 6 7

4C × 2! × 5!
2
  
(2 odd places (O,U) (consonants)
for vowels)

(consonants)

(vi) 𝐽 𝑂 𝐷 𝐻 𝑃 𝑈 𝑅

vowels

Geometric Based Selections:

If there are n points in a plane of which 𝑚(< 𝑛) are collinear, then
(a) Total number of different straight lines obtained by joining these 𝑛 points is
n
C2 −m C2 + 1
(b) Total number of different triangles formed by joining these 𝑛 points is
n
C3 −m C3
(c) Number of diagonals in polygon of 𝑛 sides is
n( n − 3)
n
C2 − n i.e.
2
(d) If 𝑚 parallel lines in a plane are intersected by a family of other 𝑛 parallel lines. Then total number of
parallelograms so formed is
mn( m − 1)( n − 1)
m
C2 n C2 i.e.
4
(e) Number of triangles formed by joining vertices of convex polygon of n sides is n C3 of which

(i) Number of triangles having exactly two sides common to the polygon = 𝑛
(ii) Number of triangles having exactly one side common to the polygon = 𝑛(𝑛 − 4)
(iii) Number of triangles having no side common to the polygon
n ( n − 4 )( n − 5)
n
C3 − n ( n − 4 ) − n =
6

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(f) In a plane there are ′𝑚′ circle & ′𝑛′ straight lines then maximum number of intersection point is
𝐿

𝐿1 𝐿2
𝑆1 𝑆2
𝑆

( n
C2  1 ) + ( m
C1 .n C1  2 ) + ( m
C2  2 )
(g) In a chase board :
98 98
(i) Total number of rectangle = 9 C2 .9 C2 =  = (36)2 = 1296
2 2
(ii) Number of square =
Square size Number of squares
1×1 8 × 8 = (8)2
2×2 7 × 7 = (7)2
3×3 (6)2
4×4 (5)2
5×5 (4)2
6×6 (3)2
7×7 (2)2
8×8 (1)2
Total = 12 + 22 + ………. + 82
n( n + 1)(2n + 1) 8(9)(17)
= = = 204
6 6
(iii) Number of rectangle which are not square = 1296 − 204 = 1092

Playing Cards:
A pack of playing cards consists of 52 cards of 4 suits, 13 in each, as shown in figure.

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Permutations and Combinations
Recognition of Cards:

Face Cards:
Face cards contain 12 cards all of 𝐾, 𝑄 and 𝐽 having designed a figure of a person.
i.e., Face cards = 4 + 4 + 4 = 12,

String Method and Gap Method

Illustration 30:
Number of words which can be made by using all the letters of the word JODHPUR.
(i) If all vowels are together.
(ii) If word ′𝑃𝑈𝑅′ always comes.
(iii) If letters ′𝑃𝑈𝑅′ always together.
(iv) If letters ′𝑃𝑈𝑅′ occurs together but ′𝐽′ never comes with ′𝑃𝑈𝑅′.
(v) If no vowels comes together.
Solution:
(i) Vowels Consonants TIE Method
O, U J, D, H, P, R

𝑂, 𝑈 𝐽 𝐷 𝐻 𝑃 𝑅

= 6! × 2!
 
Blocks (O,U)

(ii) 𝐽 𝑂 𝐷 𝐻 𝑃𝑈𝑅

5! blocks

𝐽 , 𝑂 , 𝐷 , 𝐻 , 𝑃, 𝑈, 𝑅
(iii)

= 5! × 3!

P, U, R

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(iv) 𝑂 , 𝐷 , 𝐻 , 𝑃, 𝑈, 𝑅

3
= 4! × 3! × 𝐶1 × 1!
   
Blocks (P,U,R) Place for J
(v) 𝐽 , 𝐷 , 𝐻 , 𝑃 , 𝑅 GAP Method
6
= 5! × 𝐶2 × 2!
  
Blocks place for (O, U)
(consonants) (vowels)

Combinatorial Arguments for Identities on nCr

𝑛
(1) 𝐶0 = 1
n
(2) C1 = n
n
(3) Cn = 1
n n
(4) Cn − r = Cr

Number of selections = Number of rejections  n

Cr = n Cn − r
n n
Proof : n Cr = = = n Cn − r
r n − r n − (n − r ) n − r
(5) If n Cx =n C y  x + y = n or 𝑥 = 𝑦
(6) Pr =n Cr r !
n

i.e. permutation is defined total number of combinations of object then arrangements of objects.
n
(7) n Cr = . n −1Cr −1
r
n n n −1 n
Proof : = = . n − 1 Cr −1
n − r r ( n − 1) − ( r − 1) . r r − 1 r

Illustration 31:
10 9 10.9
10
C2 = . C1 =
2 2
n +1
(8) Cr + Cr −1 = Cr
n n

Solution:
n n
Proof : L.H.S = n Cr + n Cr −1 = +
n – r· r n – r + 1· r − 1

u n
= +
n – r· r –1 ( n – r + 1) n − r· r − 1

n 1 1 
= +
n – r r − 1  r n – r + 1 
n  ( n − r + 1) + r  n +1 n +1
=   = = Cr = R.H.S.
n – r r –1  r( n − r + 1)  n − r + 1 r

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Permutations and Combinations
Illustration 32:
52
C15 +52C36 = ?
Solution:
52
C37 +52C36 =53 C37
Illustration 33:
r
Cr + r +1Cr + r +2Cr + ......... + n Cr = ?
Solution:
r
Cr + r +1Cr + r +2Cr + ......... + n Cr
= r +1
Cr +1 + r +1Cr + r +2Cr + ......... + n Cr  r Cr = r +1Cr +1 = 1
r +2
= Cr +1 + r +2Cr + ......... + nCr
r +3
= Cr +1 + r +3Cr + ......... + nCr
= n Cr +1 + n Cr
n +1
= n +1
Cr +1 = Cn−r
Illustration 34:
n
Cr n − ( r − 1)
(9) n =
Cr −1 r
Solution:
Proof :
n
Cr
LHS = n
C r −1
 n   n 
n
Cr =  =  ...(1)
 n − r r   n − r r r −1
 n   n 
n
Cr −1 =  =  ...(2)
 n − r + 1 r − 1  ( n − r + 1) n − r . r − 1 
Divide equation (1) by (2)
n
Cr n − ( r − 1)
n
=
Cr −1 r

Arrangement of Alike Objects

Permutations when all the Objects are not Distinct Objects
Suppose we have to find the number of ways of rearranging the letters of the word 𝐵𝑂𝑂𝑇. In this
case, the letters of the word are not all different. There are 2𝑂𝑠, which are of the same kind.
Let us treat, temporarily, the 2𝑂𝑠 as different, say, 𝑂1 and 𝑂2 . The number of permutations of
4 −different letters, in this case, taken all at a time is 4!. Consider one of these permutations say, 𝐵𝑂𝑂𝑇.
Corresponding to this permutation, we have 2! permutations 𝐵𝑂1 𝑂2 𝑇 and 𝐵𝑂2 𝑂1 𝑇 which will be
exactly the same permutation if 𝑂1 and 𝑂2 are not treated as different, i.e., if 𝑂1 and 𝑂2 are the
same 𝑂 at both places.

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4!
Therefore, the required number of permutations = 3  4 = 12
2!
BO1O2T 
⎯⎯
→ BOOT
BO2O1T 
TO1O2 B 
⎯⎯
→TOOB
TO2O1 B 
BO1TO2 
⎯⎯
→ BOTO
BO2TO1 
TO1 BO2 
⎯⎯
→TOBO
TO2 BO1 
BTO1O2 
⎯⎯
→ BTOO
BTO2O1 
TBO1O2 
⎯⎯
→TBOO
TBO2O1 
O1O2 BT 
⎯⎯
→ OOBT
O2O1 BT 
O1 BO2T 
⎯⎯
→ OBOT
O2 BO1T 
O1TO2 B 
⎯⎯
→ OTOB
O2TO1 B 
O1 BTO2 
⎯⎯
→ OBTO
O2 BTO1 
O1TBO2 
⎯⎯
→ OTBO
O2TBO1 
O1O2TB 
⎯⎯
→ OOTB
O2O1TB 
Let us now find the number of ways of rearranging the letters of the word 𝐼𝑁𝑆𝑇𝐼𝑇𝑈𝑇𝐸. In this case there
are 9 letters, in which I appears 2 times and 𝑇 appears 3 times. Temporarily, let us treat these letters
different and name them as 𝐼1 , 𝐼2 , 𝑇1 , 𝑇2 , 𝑇3 . The number of permutations of 9 different letters, in this case,
taken all at a time is 9!. Consider one such permutation, say, 𝐼1 𝑁𝑇1 𝑆𝐼2 𝑇2 𝑈 𝐸 𝑇3 • Here if 𝐼1 , 𝐼2 are not same
and 𝑇1 , 𝑇2 , 𝑇3 are not same, then 𝐼1 , 𝐼2 can be arranged in 2! Ways and 𝑇1 , 𝑇2 , 𝑇3 an be arranged in 3! Ways.
Therefore, 2! × 3! Permutations will be just the same permutation corresponding to this chosen
9!
permutation 𝐼1 𝑁𝑇1 𝑆𝐼2 𝑇2 𝑈𝐸𝑇3 . Hence total number of different permutations will be
2!3!
Illustration 35:
Find the number of permutations of the letters of the word 𝐴𝐿𝐿𝐴𝐻𝐴𝐵𝐴𝐷.
Solution:
Here, there are 9 objects (letters) of which there are 4𝐴’𝑠, 2𝐿’𝑠 and rest are all different.
9! 5  6  7  8  9
Therefore, the required number of arrangements = = = 7560
4!2! 2

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Permutations and Combinations
Illustration 36:
In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour
are indistinguishable?
Solution:
Total number of discs are 4 + 3 + 2 = 9. Out of 9 discs, 4 are of the first kind (red), 3 are of the second kind
(yellow) and 2 are of the third kind (green).
9!
Therefore, the number of arrangements = 1260
4!3!2!
Illustration 37:
In how many ways the letters of the word "𝐴𝑒𝑟𝑜𝑝𝑙𝑎𝑛𝑒" can be arranged without altering the relative
positions of vowels & consonants?
(A) 620 (B) 720 (C) 820 (D) 920
Ans. (B)
Solution:
The consonants in their positions can be arranged in 4! = 24 ways.
5!
The vowels in their positions can be arranged in = 30 ways
2!2!
⇒ Total number of arrangements = 24 × 30 = 720

Total Number of Combinations:

(a) Given n different objects, the number of ways of selecting atleast one of them is,
n
C1 +n C2 +n C3 + ..... +n Cn = 2n − 1 . This can also be stated as the total number of combinations of 𝑛
distinct things.
(b) (i) Total number of ways in which it is possible to make a selection by taking some or all out of
𝑝 + 𝑞 + 𝑟 + ...... things, where 𝑝 are alike of one kind, 𝑞 alike of a second kind, 𝑟 alike of third kind
& so on is given by :
( p + 1)( q + 1)( r + 1) ..... − 1
(ii) The total number of ways of selecting one or more things from 𝑝 identical things of one kind,
𝑞 identical things of second kind, 𝑟 identical things of third kind and 𝑛 different things is given by :
( p + 1)( q + 1)( r + 1) .....2  − 1
n

Illustration 38:
There are 3 books of mathematics, 4 of science and 5 of english. How many different collections can be
made such that each collection consists of-
(i) one book of each subject?
(ii) at least one book of each subject?
(iii) at least one book of English?
Solution:
(i) 3 𝐶1 × 4 𝐶1 × 5 𝐶1 = 60
(ii) (23 – 1) (24 – 1) (25 – 1) = 7 × 15 × 31 = 3255
(iii) (25 – 1) (23 ) (24 ) = 31 × 128 = 3968 Ans.

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Illustration 39:
Find the number of groups that can be made from 5 red balls, 3 green balls and 4 black balls, if at least one
ball of all colours is always to be included. Given that all balls are identical except colours.
Solution:
After selecting one ball of each colour, we have to find total number of combinations that can be made from
4 red. 2 green and 3 black balls. These will be (4 + 1)(2 + 1)(3 + 1) = 60

Divisors:
Let 𝑁 = 𝑝𝑎 . 𝑞 𝑏 . 𝑟 𝑐 ....... where 𝑝, 𝑞, 𝑟........ are distinct primes & 𝑎, 𝑏, 𝑐....... are natural numbers then :
(a) The total numbers of positive divisors of 𝑁 including 1 & 𝑁 is = (𝑎 + 1) (𝑏 + 1) (𝑐 + 1).......
(b) The sum of these divisors is
= (𝑝0 + 𝑝1 + 𝑝2 + . . . . + 𝑝𝑎 ) (𝑞0 + 𝑞1 + 𝑞 2 + . . . . + 𝑞 𝑏 ) (𝑟 0 + 𝑟1 + 𝑟 2 + . . . . +𝑟 𝑐 ). ..
(c) Number of ways in which 𝑁 can be resolved as a product of two positive factor is =
1
( a + 1) (b + 1) (c + 1)...... if N is not a perfect square
2
1
2
(a + 1) (b + 1) (c + 1)...... + 1 if N is a perfect square
(d) Number of ways in which a composite number 𝑁 can be resolved into two factors which are relatively
prime (or coprime) to each other is equal to 2𝑛−1 where n is the number of different prime factors in 𝑁.
Note :
(i) Every natural number except 1 has atleast 2 divisors. If it has exactly two divisors then it is called a
prime. System of prime numbers begin with 2. All primes except 2 are odd.
(ii) A number having more than 2 divisors is called composite. 2 is the only even number which is not
composite.
(iii) Two natural numbers are said to be relatively prime or coprime if their 𝐻𝐶𝐹 is one. For two natural
numbers to be relatively prime, it is not necessary that one or both should be prime. It is possible that
they both are composite but still coprime, eg. 4 and 25.
(iv) 1 is neither prime nor composite however it is co-prime with every other natural number.
(v) Two prime numbers are said to be twin prime numbers if their non-negative difference is 2
(e.g. 5 & 7, 19 & 17 etc).
(vi) All positive divisors except the number itself are called proper divisors.
Illustration 40:
Find the number of proper divisors of the number 38808. Also find the sum of these divisors.
Solution:
(i) The number 38808 = 23 . 32 . 72 . 11
Hence the total number of divisors (excluding itself i.e. 38808)
= (3 + 1)(2 + 1)(2 + 1)(1 + 1)– 1 = 71
(ii) The sum of these divisors
=(20 + 21 + 22 + 23 )(30 + 31 + 32 )(70 + 71 + 72 )(110 + 111 )– 38808
= (15)(13)(57)(12)– 38808 = 133380 – 38808 = 94572.

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Permutations and Combinations
Permutation & Combination of Things which are NOT all Different
Combination of Things which are not all Different :
(i) Number of ways of selection of ′𝑟′ identical things out of ′𝑛′ identical things = 1
(ii) Number of ways of selecting zero or more things out of ′𝑛′ identical things = 𝑛 + 1
Proof :
Selecting none thing = 1 way
Selecting 1 thing = 1 way
Selecting 2 things = 1 way
:
:
Selecting 𝑛 things = 1 way
Total number of ways = 1 + 1 + 1 . . . . . . (𝑛 + 1) times = 𝑛 + 1
(iii) Number of ways of selection of one or more things out of which,
′𝑝′ are alike of one kind,
′𝑞′ are alike of second kind,
′𝑟′ alike of third kind and
remaining ′𝑠′ are different is = (𝑝 + 1)(𝑞 + 1)(𝑟 + 1)2𝑆 – 1
Proof :
Selecting none thing (out of 𝑝 alike things) = 1 way
Selecting 1 thing (out of p alike things) = 1 way
Selecting 2 things (out of p alike things) = 1 way
:
:
Selecting 𝑝 things (out of 𝑝 alike things) = 1 way
Total number of ways = 1 + 1 + 1 . . . . . . . . . (𝑝 + 1) times = 𝑝 + 1
Similarly for 𝑞 alike, total ways = 𝑞 + 1
Similarly for 𝑟 alike, total ways = 𝑟 + 1
For s different things total ways of selection will be 2𝑆 , i.e. any item is selected or not.
So total number of required ways = (𝑝 + 1)(𝑞 + 1)(𝑟 + 1)2𝑆 – 1
(1 is subtracted when no item is selected)
(iv) Number of ways of selection of atleast one thing of each kind in point (iii) is = 𝑝. 𝑞. 𝑟 … = (2𝑠 − 1)
Illustration 41:
Find the number of ways in which one or more letter be selected from the letters "𝐴𝐴𝐴𝐴𝐵𝐵𝐶𝐶𝐶𝐷𝐸𝐹"
Solution:
Total number of ways = (4 + 1)(2 + 1)(3 + 1)23 – 1 = 479
𝐴 𝐵 𝐶 𝐷𝐸𝐹
Illustration 42:
How many total no. of ways of selections of letter can be possible from the word 𝑀𝐼𝑆𝑆𝐼𝑆𝑆𝐼𝑃𝑃𝐼.
OR
How many ways of selection of atleast one letter of word 𝑀𝐼𝑆𝑆𝐼𝑆𝑆𝐼𝑃𝑃𝐼.
Solution:
𝑀𝐼𝑆𝑆𝐼𝑆𝑆𝑃𝑃𝐼
𝑀1 , 𝐼 4 , 𝑆 4 , 𝑃2
   
(1 + 1)(4 + 1)(4 + 1)(2 + 1) − 1 = 2 ⋅ 5 ⋅ 5 ⋅ 3 − 1 = 149

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Illustration 43:
It is given that 4 Apples, 3 Mangoes, 2 Bananas, 2 Oranges, consider the following cases.
Case-I : Fruits of same species are alike and rests are different ,then
(i) Find the number of ways if atleast one fruit is selected.
(ii) Find the number of ways if atleast one fruit of each kind are selected.
(iii) Find the number of ways if atleast two apples & two mangoes are selected.
Case-II: Fruits of same species are different and rests are also different, then
(i) Find the number of ways, atleast one fruit is selected.
(ii) Find the number of ways, atleast one fruit of each kind is selected.
Solution:
Case-I :
(i) Apples can be selected in (4 + 1) ways
Total number of ways = (4 + 1)(3 + 1)(2 + 1)(2 + 1)– 1 = 179
(ii) Since we need atleast one fruit of each kind, Apples can be selected in 4 ways.
So total number of ways = 4 × 3 × 2 × 2 = 48
(iii) Since we need atleast two apples & two mangoes which can be selected in one way.
So total number of ways = (2 + 1)(1 + 1)(2 + 1)(2 + 1) = 54
Case-II :
(i) Since fruits of same kind are different.
Apples can be selected in 24 ways
Total number of ways = 24 × 23 × 22 × 22 – 1 = 2047
(ii) Since we need atleast one fruit of each kind, Apples can be selected in 24 – 1 ways.
So total number of ways = (24 – 1) × (23 – 1) × (22 – 1) × (22 – 1)

Formation of Groups
Number of ways in which (𝑚 + 𝑛) different things can be divided into two groups containing 𝑚 & 𝑛 things
( m + n )!
(i) If 𝑚 ≠ 𝑛, then number of ways is
m! n!
Explanation:
To find the number of ways in which (𝑚 + 𝑛) different things can be divided into two unequal groups, it
is equivalent to select ′𝑚′ persons. Since for each selection of ‘𝑚′ persons there will be a corresponding
rejection of 𝑛 persons hence each selection of 𝑚 and 𝑎 corresponding rejection of 𝑛 people will give a group.

𝑚+𝑛 ( m + n )!
 Number of groups = 𝐶𝑚 =
m! n!

Note : If these groups are to be distributed among two persons or groups are to be named, then number
( m + n )!
of ways is =  2!
m! n!
(2n )!
(ii) If 𝑚 = 𝑛, then number of ways is
n ! n ! 2!

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Explanation:
Consider 4 different toys 𝑇1 𝑇2 𝑇3 𝑇4
When 𝑇1 𝑇2 is selected and 𝑇3 𝑇4 is rejected  one way of forming the group.
4!
When 𝑇3 𝑇4 is selected and 𝑇1 𝑇2 is rejected is not a different group hence gives double answer.
2! 2!
4!
Therefore, the correct answer is .
2! 2! 2!
Hence the number of ways in which 2𝑛 different things can be divided into two equal groups
2n
Cn 2n !
= = …(i)
2! n ! n ! 2!

Note : If these groups are to be distributed among two persons or groups are to be named, then number
2n !
of ways is =  2!
n ! n ! 2!
Proof : Divide 𝑃1 , 𝑃2 , 𝑃3 , 𝑃4 in two groups
Team - A Team - B
P1P2 P3P4
P1P3 P2P4
P1P4 P2P3
P2P3 P1P4
P2P4 P1P3
P3P4 P1P2

We see that half of the case are repeated.

4!
Thus gives us wrong answer.
2! 2!
4!
Correct answer =
(2!) (2!) (2!)
Actually, counting all such cases we observe that regrouping appears when equal size groups are
required. To avoid false counting we divided by factorial of number of equal size groups.

2. Similarly (𝑚 + 𝑛 + 𝑝) different things can be divided into 3 unequal groups of 𝑚, 𝑛 and 𝑝 things is
( m + n + p )!
m! n! p!
(3n )!
(i) If the groups are all equal then the number of ways =
( n !)3 3!

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(ii) If these groups are to be distributed among 3 persons or group are to be named, then number of
(3n )!.3!
ways =
( n !)3 3!
Proofs and Explanation of Above:
To understand the article considers 10 children to be divided into three unequal groups of 2, 3 and 5.
10!
First make two groups of 2 and 8 and this can be done in way say. 𝐴𝐵/𝐶𝐷𝐸𝐹𝐺𝐻𝐼𝐽.
2! 8!
8!
Consider one such group of 8 which can be divided into two groups of 3 and 5 in ways.
3! 5!
10! 8! 10!
Hence total = . =
2! 8! 3!5! 2!3!5!
Similar explanation will be valid if initial groups in 3 and 7 and then split 7 in 2 and 5. However if 10 is
10!
divided into two groups of 5 each initially, which can be done in ways …(i)
5!5!2!
One such grouping is say
𝐴𝐵𝐶𝐷𝐸𝐹𝐺𝐻𝐼𝐽
Consider 𝐹 𝐺 𝐻 𝐼 𝐽 keeping 𝐴 𝐵 𝐶 𝐷 𝐸 as it is. Now the group 𝐹 𝐺 𝐻 𝐼 𝐽 can be divided into two groups of 2
5!
and 3 in ways and similarly when 𝐹 𝐺 𝐻 𝐼 𝐽 is kept as it is, 𝐴 𝐵 𝐶 𝐷 𝐸 can be divided into two groups
2!3!
5! 5!
of 2 and 3 in ways. Hence one group (each of 5) given by (i) generated 2. different groups of 2, 3, 5.
2!3! 2!3!
10! 2.5! 10!
 Total number of groups = . =
5!5!2! 2!3! 2!3!5!
Similarly if 𝑚 = 𝑛 = 𝑝 situation becomes different.
Consider 𝑇1 𝑇2 𝑇3 𝑇4 𝑇5 𝑇6 to be divided into 3 equal groups.
6!
When we say 6 𝐶2 . 4 𝐶2 = is totally wrong why?
2!2!2!
Selected in 6C2 Selected in 4C2 Rejected in 4C2
(A)
T1T2 T3T4 T5T6
T1T2 T5T6 T3T4
T3T4 T1T2 T5T6
T3T4 T5T6 T1T2
T5T6 T1T2 T3T4
T5T6 T3T4 T1T2
R S G
Note the all these six groups are counted in (𝐴), however they are identical. Hence the answer in (𝐴) is as
many numbers of times more as many numbers of times these equal groups can be arranged i.e. 3! times.
6!
Hence the correct number of groups is equal to .
2!2!2!3!
6!×3!
In case these 6 toys are to be distributed between 𝑅/𝑆/𝐺 then our answer will be = .
2!2!2!3!

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Permutations and Combinations
Illustration 44:
In how many ways 10 children be divided into
(i) 2 groups having 4 & 6 children.Error! Bookmark not defined.
(ii) 2 groups each having equal number of children.
(iii) 3 groups, if groups having 2, 3, & 5 children. Error! Bookmark not defined.
(iv) 3 groups, if groups having 4, 4 & 2 children.
(v) 5 groups 𝐴, 𝐵, 𝐶, 𝐷, 𝐸 having equal number of children.
Solution:

10
(i) 
6· 4

10
(ii) 
5· 5× 2

10
(iii) 
2· 3· 5

10
(iv) 
4· 4· 2× 2

 10  10
(v)    × 5=
( 2)
5
 2 · 2 · 2 · 2 · 2 × 5
Illustration 45:
Find number of ways by which 30 Jawan's can be divided into three groups of 12, 10, & 8 and send to three
different boarder's.
Solution:
(30!)×3!
Total ways =
(8!)(10!)(12!)
In above case if group are equal size (i.e., group of 10 each)
Send to three boarder's

(30!)×(3!)
=
(10!)3 (3)!

Three equal size groups

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 JEE (Main + Advanced) : Mathematics
Illustration 46:
Find number of ways by which five different objects given to three students.
Solution:
Two cases possible {1, 1, 3} {1, 2, 2}
 5! 5! 
 2
+ 2  3!
 (1!) 3!×2! 1!(2!) ×2! 
Illustration 47:
Number of ways in which 8 persons can be seated in three diff. taxies each having 3 seats for passengers
and duly numbered if
(a) If internal arrangement of persons inside the taxi is immaterial.
(b) If internal arrangement also matters
Solution:

 8! 1
(a)  2!3!3! × 2!  3!
 

 8! 1 
(b) Using grouping  ×  3! 3! 3! 3! = 9!
 2! 3! 3! 2!  
or arrange 8 people in 9 seat 9 𝐶8 × 8! = 9!
Illustration 48:
Find the number of ways of dividing 52 cards among 4 players equally such that each gets exactly one Ace.
48! 47! 46! 44!
(A) 4 × 4! (B) 4 × 5! (C) 4 × 8! (D) × 5!
(12!) (13!) (14!) (15!)4
Ans. (A)
Solution:
48!
Total number of ways of dividing 48 cards (Excluding 4 Aces) in 4 groups = 4
(12!) 4!
48!
Now, distribute exactly one Ace to each group of 12 cards. Total number of ways = 4 × 4!
(12!) 4!
48! 48!
Now, distribute these groups of cards among four players = 4 × 4! 4! = 4 × 4!
(12!) 4! (12!)

Principle of Inclusion and Exclusion

In the Venn's diagram (i), we get

𝑛(𝐴 ∪ 𝐵) = 𝑛(𝐴) + 𝑛(𝐵)– 𝑛(𝐴 ∩ 𝐵)

In the Venn's diagram (ii), we get

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Permutations and Combinations

𝑛(𝐴 ∪ 𝐵 ∪ 𝐶)
= 𝑛(𝐴) + 𝑛(𝐵) + 𝑛(𝐶)– 𝑛(𝐴 ∩ 𝐵) – 𝑛(𝐵 ∩ 𝐶) – 𝑛(𝐴 ∩ 𝐶) + 𝑛(𝐴 ∩ 𝐵 ∩ 𝐶)
Illustration 49:
How many words can be formed using all the letters of the word 𝐻𝑂𝑁𝑂𝐿𝑈𝐿𝑈 if no two alike letters are
together.
Solution:
Let 𝐴 represents ways when 𝑂𝑂 together, 𝐵 when 𝐿𝐿 together, 𝐶 when 𝑈𝑈 together.
Required ways = Total ways – [When all three alike letters together + when 2 alike letters together
+ when one alike letter together]
= Total ways – [𝑛(𝐸3 ) + 𝑛(𝐸2 ) + 𝑛(𝐸1 )] …(i)
8!
Total ways = = 5040
(2!) (2!)(2!)
(a) 𝑛(𝐸3 ) = 𝐴 ∩ 𝐵 ∩ 𝐶 (Region 7)
i.e., H N OO LL UU
𝑛(𝐸3 ) = 5! = 120
(b) 𝑛(𝐸2 ) = 3 [(𝐴 ∩ 𝐵)– (𝐴 ∩ 𝐵 ∩ 𝐶)] or [Region 4 + 5 + 6]
𝑛(𝐸3 )

 6! 
= 3  − 5!  = 720
 2! 

i.e., H OO N LL UU
(c) 𝑛(𝐸1 ) = 3 [𝐴– {(𝐴 ∩ 𝐵) + (𝐴 ∩ 𝐶)} + (𝐴 ∩ 𝐵 ∩ 𝐶)]
 7!  6!  
=  −   2  + 5! = 1980
 (2!)(2!)  2!  
𝑠𝑡
Put in 1
Required ways = 5040– [120 + 720 + 1980] = 2220 Ans.

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Alternate solution
No. of words = Total words – [At least on pair exist]
= 5040– 𝑛(𝐴 ∪ 𝐵 ∪ 𝐶)
𝑛(𝐴 ∪ 𝐵 ∪ 𝐶) = 𝑛(𝐴) + 𝑛(𝐵) + 𝑛(𝐶) − 𝑛(𝐴 ∩ 𝐵) − 𝑛(𝐵 ∩ 𝐶) − 𝑛(𝐶 ∩ 𝐴) + 𝑛(𝐴 ∩ 𝐵 ∩ 𝐶)
 7 7 7 6 6 6
  + +  − − − + 5 = (5040) – (2820) = 2220
2 2 2 2 2 2 2 2 2
Illustration 50:
In How many ways two Americans, two Britishers, two Chinese and one person each of France, Germany,
Egypt and Dutch can be sitted if persons of same nationality are to be separated.
Solution:
Number of ways = Total arrangements – [when atleast 1 pair exist together]
= 10! – 𝑛(𝐴 ∪ 𝐵 ∪ 𝐶)
 n( A) + n( B ) + n(C )– n( A  B )– 
= 10–  
 n( B  C )– n(C  A) + n( A  B  C )

( )
= 10 − ( 9  2)  3 − (( 8  2  2)  3 + 7  2  2  2 
 
= 3628800 – [2177280 – 483840 + 40320]
= 3628800 – [1733760] = 1895040

Circular Permutation
Permutation of objects in a row is called as linear permutation. If we arrange the objects along a closed
curve it is called as circular permutation.
Thus in, circular permutation, we consider one object fixed and the remaining objects are arranged as in
the case of a linear arrangements.
Case-I : When object are different:
Theorm-1: The number of circular permutation of 𝑛 distinct objects is (𝑛 − 1)!
Proof : Consider 5 objects 𝐴, 𝐵, 𝐶, 𝐷, 𝐸 to be arranged around a closed curve is called circular permutation.

All are same

Let the total number of circular permutation be 𝑥. Above circular permutation is equivalent to 5 linear
permutations given by 𝐴𝐵𝐶𝐷𝐸𝐹, 𝐸𝐴𝐵𝐶𝐷, 𝐷𝐸𝐴𝐵𝐶, 𝐶𝐷𝐸𝐴𝐵, 𝐵𝐶𝐷𝐸𝐴 that is one circular permutation is
equivalent to 5𝑥 linear permutation given by
𝑥 .5 = 5 !
5! 5.(5 − 1)!
𝑥= = = (5 − 1)!
5 5
Similarly for 𝑛 objects 𝑛𝑥 = 𝑛!
n!
x= = ( n − 1)!
n
(i) 𝑛 distinct things taken all at a time and arranged along circle in (𝑛 – 1)! ways
(ii) Taken 𝑟 things out of 𝑛 distinct things at a time and arranged along circle in 𝑛 𝐶𝑟 . (𝑟– 1)! ways.
Note : In the above theorem anti-clockwise and clockwise order of arrangements are considered as
distinct permutations.

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Permutations and Combinations
Illustration 51:
Find the number of ways in which 10 children can sit in a mary go round relative to one another.
Solution:
Here clockwise and anticlockwise arrangements are different.
Thus required ways = (10 − 1)! = 9!
Theorm-2 : If anticlockwise and clockwise are considered to be same total number of circular permutation
( n − 1)!
given by .
2
If we arrange flowers or garland beads in a necklace then there is no distinction between clockwise
& anticlockwise direction.
Turn over
Flip to right

=
Clock wise Counter clockwise Both same

Note : If we have 𝑛 different things taken 𝑟 at a time in form of a garland or necklace.

n
Cr .( r − 1)!
Required number of arrangements =
2
Illustration 52:
In how many ways garlands can be formed out of 10 different flowers, if each garland consist of :
(i) 10 flowers(ii) 6 flowers
Solution:
(i) Here clockwise and anticlockwise permutations are same
(10 − 1)! 9!
Hence total ways = =
2 2
(6 − 1)! 10 5!
(ii) Hence total ways = 10 C6  = C6 
2 2
Important note :
The distinction between clockwise and anticlockwise is ignored when a number of people have to be seated
around a table so as not to have the same neighbours.
Illustration 53:
Find the number of ways in which 9 people can be seated on a round table so that all shall not have the
same neighbours in any 2 arrangements.
Solution:
For same neighbour, clockwise and anticlockwise arrangements are same.
So total number of ways will be arrangement of 9 people taken clockwise and anticlockwise same and
8!
equal to .
2

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Distribution of Alike Objects
TYPE-1: Total number of ways in which ′𝑛′ identical coins can be distributed among ′𝑟′ persons so that
𝑛+𝑟–1 ( n + r − 1)!
each person may get any number of coin is 𝐶𝑟–1 =
( r − 1)! ( n )!
Proof: Let 6 identical coins can be distributed among 3 persons 𝑅|𝑆|𝐺

(Separators)

Out of eight places if we8 select any two, then 3

partitions are formed by 𝐶2 ways

Illustration 54:
In how many ways 10 identical coins can be distributed among four persons (beggars) if each beggar can get?
(i) any number of coins.
(ii) at least one coin.
Solution:
𝑛 = 10
𝑟=4
𝑛+𝑟–1
(i) Number of ways = 𝐶𝑟–1
10+4–1
= 𝐶4–1
13
= 𝐶3 = 286
B1 B2 B3 B4
(ii)
1 1 1 1
Give one coin to each beggars
 Remaining coins = 10 − 4 = 6
𝑛 =6&𝑟 =4
6+4–1
 Number of ways = 𝐶4–1 = 9 𝐶3
Illustration 55:
Number of ways in which 5 identical balls can be kept into 3 different boxes so that no box remains empty
will be
(A) 1 (B) 3 (C) 6 (D) 15
Ans. (C)
Solution:
Keep one ball to each box
 Remaining balls = 5 – 3 = 2
𝑛+𝑟–1 2+3–1 4 4.3
The required number of ways = 𝐶𝑟–1 = 𝐶3–1 = 𝐶2 = =6
2

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Permutations and Combinations
Problems Based on Integral Solutions
Type-2 : Number of non-negative integral solutions of an equation 𝑥1 + 𝑥2 + 𝑥3 . . . . . . . + 𝑥𝑟 = 𝑛 is
= 𝑛+𝑟–1 𝐶𝑟–1
Illustration 56:
For an equation 𝑥 + 𝑦 + 𝑧 = 20, then find
(i) No. of non – 𝑣𝑒 integral solution.
(ii) No. of integral solution if 𝑥 > – 1, 𝑦 ≥ 0, 𝑧 > 3
(iii) No. of non – 𝑣𝑒 even integral solution.
(iv) No. of non – 𝑣𝑒 odd integral solution.
𝑥 + 𝑦 + 𝑧 = 20
Solution:
(i) 𝑥 ≥ 0𝑦 ≥ 0𝑧  0
𝑛 = 20𝑟 = 3
𝑛+𝑟–1 22
 Number of solutions = 𝐶𝑟–1 = 𝐶2
(ii) 𝑥 + 𝑦 + 𝑧 = 20
𝑥 > –1 𝑦 ≥ 0 𝑧 > 3
𝑥 ≥ 0𝑦 ≥ 0 𝑧 ≥ 4
𝑁 = 20– 4 = 16
 Number of solutions = 𝑛+𝑟–1 𝐶𝑟–1 = 16+3–1
𝐶3–1 = 18
𝐶2
(iii) 𝑥 + 𝑦 + 𝑧 = 20
𝑥 = 2𝑝 ; 𝑝 = 0,1,2. . . . . . . ; 𝑝 ≥ 0
𝑦 = 2𝑞 ; 𝑞 = 0,1,2. . . . . . . ; 𝑞 ≥ 0
𝑧 = 2𝑟 ; 𝑟 = 0,1,2. . . . . . . ; 𝑟 ≥ 0
𝑥 + 𝑦 + 𝑧 = 2𝑝 + 2𝑞 + 2𝑟
20 = 2(𝑝 + 𝑞 + 𝑟)
𝑝 + 𝑞 + 𝑟 = 10
 Number of solutions = 𝑛+𝑟–1 𝐶𝑟–1 = 10+3–1
𝐶3–1 = 12
𝐶2
(iv) 𝑥 + 𝑦 + 𝑧 = 20
Not possible
Illustration 57:
𝑥 + 𝑦 + 𝑧 = 21 ; Non negative odd.
Solution:
𝑥 + 𝑦 + 𝑧 = 21
𝑥 = 2𝑝 + 1 ; 𝑝 ≥ 0
𝑦 = 2𝑞 + 1 ; 𝑞 ≥ 0
𝑧 = 2𝑟 + 1 ; 𝑟 ≥ 0
𝑥 + 𝑦 + 𝑧 = 2𝑝 + 2𝑞 + 2𝑟 + 3
21 = 2(𝑝 + 𝑞 + 𝑟) + 3
𝑝+𝑞+𝑟 = 9
𝑛 = 9𝑟 = 3
𝑛+𝑟–1 11
 Number of solutions = 𝐶𝑟–1 = 𝐶2

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Some More Combinatorial Problems
Illustration 58:
Number of ways in which 8 people can be arranged in a line if 𝐴 and 𝐵 must be next each other and 𝐶 must
be somewhere behind 𝐷, is equal to
(A) 10080 (B) 5040 (C) 5050 (D) 10100
Ans. (B)
Solution:
𝐴𝐵𝐶 𝐷𝐸 𝐹 𝐺 𝐻
7 2
= 5040
2
𝐴 & 𝐵 are tied with string
So there are total 7 units.
Which can be arranged by 7× 2 ways

(A & B can beinterchanged)
Now 𝐶 is somewhere being 𝐷
 We divide it by 2
Hence option (B) is correct.
Illustration 59:
Number of ways in which ′𝑚′ different toys can be distributed in ′𝑛′ children if every child may receive any
number of toys, is
(A) 𝑛𝑚 (B) 𝑚 𝐶𝑛 (C) 𝑛 𝐶𝑚 (D) 𝑚𝑛
Ans. (A)
Solution:
Number of ways in which ′𝑚′ different toys can be distributed in ′𝑛′ children if every child may receive any
number of toys = 𝑛𝑚 .
Illustration 60:
Six married couple are sitting in a room. Find the number of ways in which 4 people can be selected so that
they do not form a couple.
(A) 240 (B) 280 (C) 255 (D) 480
Ans. (A)
Solution:
6
𝐶4 . 2 𝐶1 . 2 𝐶1 . 2 𝐶1 . 2 𝐶1 = 240
Illustration 61:
Number of 5 digit numbers divisible by 25 that can be formed using only the digits 1, 2, 3, 4, 5 & 0 taken
five at a time is
(A) 2 (B) 32 (C) 42 (D) 52
Ans. (C)
Solution:
If the number is divisible by 25 and digit used to form the number are 1, 2, 3, 4, 5, 0 the can last 2 digit can
be 25 or 50.

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Permutations and Combinations
Case-I Case-II
2 5 5 0

3 3 2 4 3 2
(1, 3, 4) (1, 2, 3, 4)

=3×3×2= 4×3×2
= 18 = 24
Total No. = 18 + 24 = 42
Station Problems
Illustration 62:
There are 𝑛 intermediate stations on a railway line from one terminus to another. In how many ways can
the train stop at 3 of these intermediate stations if
(a) all the three stations are consecutive
(b) at least two of the stations are consecutive
(c) no two of these stations are consecutive.
Solution:
(a) The number of triples of consecutive stations, viz.
𝑆1 𝑆2 𝑆3 , 𝑆2 𝑆3 𝑆4 , 𝑆3 𝑆4 𝑆5 , . . . . . . . . 𝑆𝑛–2 𝑆𝑛–1 𝑆𝑛
(b) The total number of consecutive pair of stations, viz.
𝑆1 𝑆2 , 𝑆2 𝑆3 . . . . . . . . . 𝑆𝑛–1 𝑆𝑛
is (𝑛 – 1).
Each of the above pair can be associated with a third station in (𝑛– 2) ways. Thus, choosing a pair of stations
and any third station can be done in (𝑛– 1) (𝑛– 2) ways. The above count also includes the case of three
consecutive stations. However, we can see that each such case has counted twice. For example, the pair
𝑆4 𝑆5 combined with 𝑆6 and the pair 𝑆5 𝑆6 combined with 𝑆4 are identical.
Hence, subtracting the excess counting, the number of ways which three stations can be chosen so that at
least two of them are consecutive
n( n − 1)( n − 2)
= (𝑛– 1) (𝑛– 2)– (𝑛– 2) = (𝑛– 2)2 .
1.2.3
(c) Without restriction, the train can stop at any three stations in 𝑛 𝐶3 ways.
Hence, the number of ways the train can stop so that no two stations are consecutive
= 𝑛 𝐶3 – (𝑛– 2)2 = – (𝑛– 2)2
 n2 − n − 6n + 12  ( n − 2)( n − 3)( n − 4) n −2
= (𝑛 – 2)  = = C3
 6  6

Summation of Numbers
Illustration 63:
Find sum of all the numbers greater then 10000 formed by the digits 1,3,5,7,9 if no digit being repeated.
Solution:
All possible numbers = 5! = 120
If one occupies the units place then total numbers = 24.

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 JEE (Main + Advanced) : Mathematics

Hence 1 enjoys units place 24 times 1

|||ly 1 enjoys each place 24 times
Sum due to 1 = 1 × 24 (1 + 10 + 102 + 103 + 104 )
|||ly Sum due to the
digit 3 = 3 × 24 (1 + 10 + 102 + 103 + 104 )
: : : : : : :
Required total sum = 24 (1 + 10 + 10 + 103 + 104 ) (1 + 3 + 5 + 7 + 9)
2

Illustration 64:
Find sum of all the numbers greater than 10000 formed by the digit 0, 1, 2, 4, 5 no digit being repeated.
Solution:
Using all the given digits we can form a five digit number except when zero is at first place.
So to find the sum of all the possible five digit number
= (Sum of all possible arrangement) – (Sum of all the arrangements when zero is at first place)
5!
 5 different digits can be arranged in 5! ways so each digit will appear at every place = times
5
i.e. 24 times
Sum of all digits at unit place = 24(0 + 1 + 2 + 4 + 5)
Sum of all digits at ten's place = 24(0 + 1 + 2 + 4 + 5)
........................................................................
Sum of all digits at 10000𝑡ℎ place = 24(0 + 1 + 2 + 4 + 5)
In this way sum of all possible arrangement = 24(0 + 1 + 2 + 4 + 5) [1 + 10 + 102 + 103 + 104 ]
when zero is at first place 4 digit number will be formed.
Each number will appear 6 times at every place.
Sum of all 4 digit number at unit place = 6 (1 + 2 + 4 + 5)
Sum of all 4 digit number at ten's place= 6 (1 + 2 + 4 + 5)
........................................................................
Hence sum of all four digit numbers = 6 (1 + 2 + 4 + 5) (1 + 10 + 102 + 103 )
Required sum = 24[0 + 1 + 2 + 3 + 4 + 5] [1 + 10 + 102 + 103 + 104 ]
– 6(1 + 2 + 4 + 5) (1 + 10 + 102 + 103 )
Illustration 65:
Find the sum of the five digit numbers that can be formed using the digits 3, 4, 5, 6, 7 not using any digit
more than once in any number.
Ans. (6666600)
Solution:
If 3 is placed at units place, the remaining 4 places can be filled in 4! = 24 ways
Thus, 3 occurs at unit place 24 times.
The other digits similarly, each occurs at the unit places 24 times.
Similarly, each of the digit occurs at the other places tens, hundreds and so on, 24 times.
Hence, the required sum, is
= 24 (3 + 4 + 5 + 6 + 7) (100 + 101 + 102 + 103 + 104 )
= 24 × 25 × 11111
= 6666600

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Permutations and Combinations
Grid Problem:
Illustration 66:
Complete cartesian plane is partitioned by drawing line | | to 𝑥 and 𝑦 −axis equidistant apart like the lines
on a chess board, then the number of ways in which an ant can reach from (1, 1) to (4, 5) via shortest path.
Solution:
What ever may be the mode of travel of the ant; it has to traverse 3𝐻 (Horizontal) and 4𝑉 (Vertical) paths.
7! 7
Hence required number of ways = = C3
4! 3!

(4,5)

(1,1)

Note: If there are 𝑛 vertical and 𝑚 horizontal lines then there will be (𝑛– 1) horizontal and (𝑚– 1) vertical
paths
Illustration 67:

6
𝐵 (4,5)
5
4
(3,3)
3
𝐶
2

(1,1) 𝐴
2 3 4 5 6

Number of ways in which an insect can reach from (1, 1) to (4, 5) via 𝐶 but having shortest path.
Solution:
(𝐴 𝑡𝑜 𝐶) and (𝐶 𝑡𝑜 𝐵)
(1, 1) – (3, 3) and (3, 3) – (4, 5)
(𝑋 3 – 1 , 𝑌 3 – 1 ) and (𝑋 4 – 3 , 𝑌 5 – 3 )
(𝑋 2 , 𝑌 2 ) and (𝑋1 , 𝑌 2 )
Number of ways = 4 𝐶2 × 3 𝐶1 = 6 × 3 = 18

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 JEE (Main + Advanced) : Mathematics
Problems Based on Sets and Subsets
Illustration 68:
Set 𝐴 contains 4 different elements and 𝐵  𝐴. In how many ways set 𝐵 can be formed?
Solution:
Let 𝑎  𝐴 then
𝑎𝐵 ...(i)
𝑎𝐵 …(ii)
for each element of 𝐴 we have 2 options for that element to be present or absent from 𝐵.
So total cases = 2 × 2 × 2 × 2 = 24
𝐵 can be formal in 24 ways.
Illustration 69:
Set 𝐴 = {1, 2, 3, 4, 5, 6} and 𝐵  𝐴 & 𝐶  𝐴. In how many ways sets 𝐵 & 𝐶 can be formed such that
𝐵𝐶=
(A) 35 (B) 36 (C) 26 (D) 35 − 1
Ans. (B)
Solution:
Let 𝑎  𝐴 then total 4 cases are possible
𝑎  𝐵, 𝑎  𝐶 ...(i)
𝑎  𝐵, 𝑎  𝐶 …(ii)
𝑎  𝐵, 𝑎  𝐶 …(iii)
𝑎  𝐵, 𝑎  𝐶 …(iv)
we require 𝐵  𝐶 = 
thus, for a, equation (ii) (iii) and (iv)
can satisfy 𝐵  𝐶 = 
therefore 𝐵  𝐶 =  ℎ𝑎𝑠 3 options
thus, No. of equation possible for 6 elements = 36
Illustration 70:
Set 𝐴 = {1, 2, 3, 4, 5, 6, 7, 8} and 𝐵  𝐴 & 𝐶  𝐴. In how many ways set 𝐵 & 𝐶 can be formed such that
𝐵  𝐶 = {2, 5, 7} ?
Solution:
1 B 1 C 

1  B 1  C  3 options 2 B 2C1 option
1 B 1 C  

3 B 3 C  4 B 4C 
 
3  B 3  C  3 options 4  B 4  C  3 options
3  B 3  C  4  B 4  C 
6 B 6C 

5 B 5C1 option 6  B 6  C  3 options
6  B 6  C 
8 B 8C 

7 B 7C1 option 8  B 8  C  3 options
8  B 8  C 
So total possible ways = 3 × 1 × 3 × 3 × 1 × 3 × 1 × 3 = 35

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Permutations and Combinations
Illustration 71:
Let 𝑋 = {1, 2, 3, …….., 10}. Find the number of pair {𝐴, 𝐵} such that 𝐴  𝑋, 𝐵  𝑋, 𝐴  𝐵 and
𝐴  𝐵 = {5, 7, 8}.
Solution:
Let 𝐴𝐵 = 𝑌, 𝐵\𝑎 = 𝑀, 𝐴\𝐵 = 𝑁 and 𝑥\𝑦 = 𝐿. then 𝑋 is the disjoint union of 𝑀, 𝑁, 𝐿 and 𝐴  𝐵. Now
𝐴  𝐵 = {5, 7, 8} is fixed. The remaining seven elements 1, 2, 3, 4, 6, 9, 10 can be distributed in any of the
remaining sets 𝑀, 𝑁, 𝐿. this can be done in 37 ways. Of these if all the elements are in the set 𝐿, then
𝐴 = 𝐵 = {5, 7, 8} and this case has to be omitted. Hence the total number of pairs {𝐴, 𝐵} such that 𝐴  𝑋,
𝐵  𝑋, 𝐴  𝐵 and 𝐴  𝐵 = {5, 7, 8}is 37 − 1.
Illustration 72:
Let 𝑆 = {1, 2, 3, 4, 5}. The total number of unordered pairs of disjoint subsets of 𝑆 is equal to –
(A) 121 (B) 122 (C) 123 (D) 124
Ans. (B)
Solution:
𝑆 = {1,2,3,4,5}
According to the definition of disjoint sets, if there exist two sets 𝐴 and 𝐵, 𝐴  𝐵 = .
Every element of 𝑆 can be an element of 𝐴 or 𝐵 or of neither of the subsets.
There exist 3 possibilities for each element
Since there are for 5 elements, there are 35 possibilities are present
Total number of ordered pairs of subsets = 35 + 1 = 244
Total number of unordered pairs = 244/2 = 122
Illustration 73:
𝐴 is a set containing 𝑛 elements. A subset 𝑃 of 𝐴 is chosen. The set 𝐴 is reconstructed by replacing the
elements of 𝑃. A subset 𝑄 of 𝐴 is again chosen. The number of ways of choosing 𝑃 and 𝑄 so that 𝑃  𝑄 = 
is :-
(A) 22𝑛 − 2𝑛 𝐶𝑛 (B) 2𝑛 (C) 2𝑛 − 1 (D) 3𝑛
Ans. (D)
Solution:
Let 𝐴 = {𝑎1 , 𝑎2 , 𝑎3 , … . . 𝑎𝑛 }. For 𝑎𝑖  𝐴, we have the following choices:
(i) 𝑎𝑖  𝑃 and 𝑎𝑖  𝑄 (ii) 𝑎𝑖  𝑃 and 𝑎𝑖  𝑄
(iii) 𝑎𝑖  𝑃 and 𝑎𝑖  𝑄 (iv) 𝑎𝑖  𝑃 and 𝑎𝑖  𝑄
Out of these only (ii), (iii) and (iv) imply 𝑎𝑖  𝑃  𝑄. Therefore, the number of ways in which none of
𝑎1 , 𝑎2 , ….𝑎𝑛 belong to 𝑃  𝑄 is 3𝑛 .

Derangement Theorem
Derangement means arrangement in which none can occupy its own place.
(i) If 𝑛 things are arranged in a row, the number of a ways they can be deranged so that 𝑟 things occupy
wrong places while (𝑛 – 𝑟) things occupy their original places, is
= 𝑛 𝐶𝑛−𝑟 𝐷𝑟
 1 1 1 1
where 𝐷𝑟 = r !  1 − + − + ..... + ( −1)r 
 1! 2! 3! r! 

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 JEE (Main + Advanced) : Mathematics
(ii) If 𝑛 things are arranged in a row, the number of ways they can be deranged so that none of them
occupies its original place, is
𝑛  1 1 1 1
= 𝐶0 𝐷𝑛 = n !  1 − + − + ..... + ( −1)n 
 1! 2! 3! n! 
Alter :
𝐷𝑛 = ways in which 𝑛 things are arranged so that all 𝑛 things occupy wrong places
= arranged without restriction
– ways in which 1 thing is in correct position while (𝑛 – 1) things are deranged
– ways in which 2 things are in correct position while (𝑛– 2) things are deranged
......................................
......................................
– ways in which all 𝑛 things are in correct position and there is no derangement
𝑛 𝑛 𝑛
= 𝑛! − 𝐶1 𝐷𝑛−1 − 𝐶2 𝐷𝑛−2 − ………. − 𝐶𝑛 𝐷0
n
= 𝑛! −  n Cr Dn − r
r =1

Thus, we have
𝐷0 = 0! = 1
𝐷1 = 1! − 1 𝐶1 𝐷0 = 0
𝐷2 = 2! − 2 𝐶1 𝐷1 − 2 𝐶2 𝐷0 = 1
𝐷3 = 3! − 3 𝐶1 𝐷2 − 3 𝐶2 𝐷1 − 3 𝐶3 𝐷0 = 2
𝐷4 = 4! − 4 𝐶1 𝐷3 − 4 𝐶2 𝐷2 − 4 𝐶3 𝐷1 − 4 𝐶4 𝐷0 = 9
𝐷5 = 5! − 5 𝐶1 𝐷4 − 5 𝐶2 𝐷3 − 5 𝐶3 𝐷2 − 5 𝐶4 𝐷1 − 5 𝐶5 𝐷0 = 44
and so on.
Short Trick :
Number of derangement of 𝑛-different objects =
1 1 1 1 1 1
Dn = n !  − + − + + ....... + (−1)n 
 0! 1! 2! 3! 4! n! 
1 1 1 1 1 1 
D2 = 2!  − +  = 2  − +  = 1
 0! 1! 2!  1 1 2 
1 1 1 1 1 1  3 − 1 
D3 = 3!  − + −  = 6  −  = 6  =2
 0! 1! 2! 3!  2 6   6 
1 1 1 1 1 1 1 1   12 − 4 + 1 
D4 = 4!  − + − +  = 24  − +  = 24  =9
 0! 1! 2! 3! 4!   2 6 24   24 
1 1 1 1 1 1 1 1 1 1   60 − 20 + 5 − 1 
D5 = 5!  − + − + −  = 120  − + −  = 120   = 44
 0! 1! 2! 3! 4! 5!   2 6 24 120   120
Illustration 74:
In how many ways three letters be posted in three addressed envelopes if
(i) all are at right place.
(ii) exactly two of them are at right place.
(iii) exactly one is at right place.
(iv) No one letter is at right place.

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Permutations and Combinations
Solution:
(i) All 3 are at correct place = 1 way
(ii) Zero way
(iii) 3 𝐶1 × 1 = 3 ways
(iv) Number of ways in which no one at right place is = 𝐾(let)
𝐾 = Total ways – (case (i) + (ii) + (iii))
𝐾 = 3! − (1 + 0 + 3)
𝐾 =6−4
𝐾 = 2 ways
𝐷2 = 1; 𝐷3 = 2; 𝐷4 = 9; 𝐷5 = 44
Short Trick :
Number of derangement of 𝑛 −different objects = 𝐷3 = 2
Illustration 75:
A person writes letters to five friends and addresses the corresponding envelopes. In how many ways can
the letters be placed in the envelops so that
(a) all letters are in the wrong envelopes.
(b) at least three of them are in the wrong envelopes.
Solution:
 1 1 1 1 1  5! 5! 5! 5!
(a) Required number of ways = 5!  1 − + − + −  = − + −
 1! 2! 3! 4! 5!  2! 3! 4! 5!
(b) Required number of ways
n
= 
r =1
n
Cr Dn − r where 𝑛 = 5

= 5 𝐶2 𝐷3 + 5 𝐶1 𝐷4 + 5
𝐶0 𝐷5
 1 1 1  1 1 1 1  1 1 1 1 1
= 10 × 3!  1 − + −  + 5 × 4!  1 − + − +  + 1 × 5!  1 − + − + − 
 1! 2! 3!   1! 2! 3! 4!   1! 2! 3! 4! 5! 
= 10(3 − 1) + 5(12 − 4 + 1) + (60 − 20 + 5 − 1) = 20 + 45 + 44 = 109.

Illustration 76:
A family consists of a grandfather, 𝑚 sons and daughters and 2𝑛 grand children. They are to be seated in a
row for dinner. The grand children wish to occupy the 𝑛 seats at each end and the grandfather refuses to
have a grand children on either side of him. In how many ways can the family be made to sit.
(A) (2𝑛)! 𝑚! (𝑚 − 1) (B) (2𝑛)! 𝑚! 𝑚
(C) (2𝑛)! (𝑚 – 1)! (𝑚 − 1) (D) (2𝑛 − 1)! 𝑚! (𝑚 − 1)
Ans. (A)
Solution:
First we select 𝑛 grand children from 2𝑛 grand children is 2𝑛 𝐶𝑛
Now arrangement of both group is 𝑛! × 𝑛!
Now Rest all (𝑚 + 1) place where we occupy the grandfather and 𝑚 sons but grandfather refuse the sit to
either side of grand children so the out of 𝑚– 1 seat one seat can be selected
Now required number of sitting in 2𝑛 𝐶𝑛 × 𝑛! × 𝑛! × (𝑚 –1) 𝐶1 . 𝑚!
12n
= × 𝑛! × 𝑛! × (𝑚 –1) 𝐶1 . 𝑚! = 2𝑛! . 𝑚! . (𝑚 − 1)
n ! n !

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 JEE (Main + Advanced) : Mathematics
Illustration 77:
′𝑛′ digits positive integers formed such that each digit is 1, 2 or 3. How many of these contain all three of
the digits 1, 2 and 3 atleast once?
(A) 3(𝑛 − 1) (B) 3𝑛 − 2.2𝑛 + 3 (C) 3𝑛 − 3.2𝑛 – 3 (D) 3𝑛 − 3.2𝑛 + 3
Ans. (D)
Solution:
Total 𝑛 −digit numbers using 1, 2 or 3 = 3𝑛
total 𝑛 −digit numbers using any two digits out of 1, 2 or 3= 3 𝐶2 × 2𝑛 − 6 = 3 × 2𝑛 − 6
total 𝑛 −digit numbers using only one digit of 1, 2 or 3 = 3
 the numbers containing all three of the digits
1, 2 and 3 at least once = 3𝑛 − (3 × 2𝑛 − 6) − 3 = 3𝑛 − 3 . 2𝑛 + 3
Illustration 78:
There are ′𝑛′ straight line in a plane, no two of which are parallel and no three pass through the same point.
Their points of intersection are joined. Then the maximum number of fresh lines thus introduced is
1 1
(A) 𝑛(𝑛 − 1)2 (𝑛 − 3) (B) 𝑛(𝑛 − 1) (𝑛 + 2) (𝑛 − 3)
12 8
1 1
(C) 𝑛(𝑛 − 1) (𝑛 − 2) (𝑛 − 3) (D) 𝑛(𝑛 + 1) (𝑛 + 2) (𝑛 − 3)
8 8
Ans. (C)
Solution:
If ′𝑛′ straight line intersect each other then total
n( n − 1)
number of intersection point is 𝑛 𝐶2 =
2
n( n −1)
Now, from these 𝑛 𝐶2 points we can make 2
C2
n −1
lines. (total old + new lines) and number of old lines are C2 × 𝑛
n( n −1)
n −1 1
So fresh lines are 2
C2 – C2 × 𝑛 = 𝑛(𝑛 − 1) (𝑛 − 2) (𝑛 − 3)
8

Illustration 79:
In how many ways can a pack of 52 cards be divided
(i) equally in four sets
(ii) equally among four players
Solution:
52 1  52! 1
(i) 4
 (ii)  4
   4!
(13!) 4!  (13!) 4! 
Illustration 80:
In how many ways 52 playing cards can be distributed among 4 players if each get equal no. of cards :
(i) If no condition
(ii) If each player can get 𝐴, 𝐽, 𝑄, 𝐾 of same suit.

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Permutations and Combinations
Solution:
52

13  52  52
(i) 13 13 13   4 
  13· 13· 13· 13· 4  ( 13)
4

36

9
(ii) 9 9 9 52 – 16 = 36(𝐴, 𝐽, 𝑄, 𝐾 of 4 suits)

  36  
 4   4  4 Remaining 𝐴, 𝐽, 𝑄, 𝐾 of same suits to 4 players.
  ( 9) · 4  
Illustration 81:
In how many ways can a pack of 52 cards be
(i) distributed among four players having 10, 12, 14 and 16 cards.
(ii) divided into four sets of 7, 15, 15 and 15 cards.
Solution:
52!
(i)  4!
10! 12! 14! 16!
52! 1
(ii) 3

7! (15!) 3!
Illustration 82:
𝑋 = {1, 2, 3, 4, . . . . . . 2017} and 𝐴 ⊂ 𝑋 ; 𝐵 ⊂ 𝑋 ; 𝐴 ∪ 𝐵 ⊂ 𝑋 here 𝑃 ⊂ 𝑄 denotes that 𝑃 is subset of
𝑄(𝑃 ≠ 𝑄). Then number of ways of selecting unordered pair of sets 𝐴 and 𝐵 such that 𝐴 ∪ 𝐵 ⊂ 𝑋.
(42017 − 32017 ) + (22017 − 1) (42017 − 32017 )
(A) (B)
2 2
42017 − 32017 + 22017
(C) (D) None of these
2
Ans. (A)
Solution:
Ordered pair = total – (𝐴 ∪ 𝐵 = 𝑋) = 4𝑛 – 3𝑛
Subsets of 𝑋 = 2𝑛 will not repeat in both but here the whole set 𝑋 has not been taken
So subsets of 𝑥 which are not repeated (2𝑛 – 1)
(4n − 3n ) − (2n − 1)
Hence unordered pair = +(2𝑛 − 1)
2
Illustration 83:
The number of ways in which 15 identical apples & 10 identical oranges can be distributed among three
persons, each receiving none, one or more is:
(A) 5670 (B) 7200 (C) 8976 (D) 7296
Ans. (C)

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 JEE (Main + Advanced) : Mathematics
Solution:
Using multinomial theorem
15 + 3 – 1 10+3–1 17 12 17  16 12  11
Total no. of ways = 𝐶15 × 𝐶10 = 𝐶15 × 𝐶10 =  = 8976
2 2
Illustration 84:
If 𝑛 identical dice are rolled, then number of possible outcomes are.
6n
(A) 6𝑛 (B) (C) ( n + 5) c5 (D) None of these
n!
Ans. (C)
Solution:
Let 𝑖 appears on 𝑎𝑖 dice 𝑖 = 1, 2, 3, 4, 5, 6
Do number of outcomes is equal to no. of solution of 𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5 + 𝑎6 = 𝑛 = ( n + 5) c5
Illustration 85:
Find the number of solutions of the equation 𝑥𝑦𝑧 = 360 when 𝑥, 𝑦, 𝑧 ∈ 𝑁
(A) 150 (B) 180 (C) 210 (D) 240
Ans. (B)
Solution:
𝑥𝑦𝑧 = 360 = 23 × 32 × 5 (𝑥, 𝑦, 𝑧 ∈ 𝑁)
𝑥 = 2𝑎1 3𝑎2 5𝑎3 (where 0  𝑎1  3, 0  𝑎2  2, 0  𝑎3  1)
𝑦 = 2𝑏1 3𝑏2 5𝑏3 (where 0  𝑏1  3, 0  𝑏2  2, 0  𝑏3  1)
𝑧 = 2𝑐1 3𝑐2 5𝑐3 (where 0  𝑐1  3, 0  𝑐2  2, 0  𝑐3  1)
 2𝑎1 3𝑎2 5𝑎3 . 2𝑏1 3𝑏2 5𝑏3 . 2𝑐1 3𝑐2 5𝑐3 = 23 × 32 × 51
 2𝑎1 +𝑏1 +𝑐1 . 3𝑎2 +𝑏2 +𝑐2 . 5𝑎3 +𝑏3 +𝑐3 = 23 × 33 × 51
 𝑎1 + 𝑏1 + 𝑐1 = 3 → 5 𝐶2 = 10
𝑎2 + 𝑏2 + 𝑐2 = 2 → 4 𝐶2 = 6
𝑎3 + 𝑏3 + 𝑐3 = 1 → 3 𝐶2 = 3
Total solutions = 10 × 6 × 3 = 180.
Illustration 86:
Find the number of solutions of the equation 𝑥𝑦𝑧 = 360 when 𝑥, 𝑦, 𝑧 ∈ 𝐼
(A) 410 (B) 520 (C) 610 (D) 720
Ans. (D)
Solution:
𝑥𝑦𝑧 = 360 = 23 × 32 × 5 (if 𝑥, 𝑦, 𝑧  𝑁)
𝑥 = 2𝑎1 3𝑎2 5𝑎3 (where 0  𝑎1  3, 0  𝑎2  2, 0  𝑎3  1)
𝑦 = 2𝑏1 3𝑏2 5𝑏3 (where 0  𝑏1  3, 0  𝑏2  2, 0  𝑏3  1)
𝑧 = 2𝑐1 3𝑐2 5𝑐3 (where 0  𝑐1  3, 0  𝑐2  2, 0  𝑐3  1)
 2𝑎1 3𝑎2 5𝑎3 . 2𝑏1 3𝑏2 5𝑏3 . 2𝑐1 3𝑐2 5𝑐3 = 23 × 32 × 51
 2𝑎1 +𝑏1 +𝑐1 . 3𝑎2 +𝑏2 +𝑐2 . 5𝑎3 +𝑏3 +𝑐3 = 23 × 33 × 51
 𝑎1 + 𝑏1 + 𝑐1 = 3 → 5 𝐶2 = 10
𝑎2 + 𝑏2 + 𝑐2 = 2 → 4 𝐶2 = 6
𝑎3 + 𝑏3 + 𝑐3 = 1 → 3 𝐶2 = 3
Total solutions = 10 × 6 × 3 = 180.
As in this question 𝑥, 𝑦, 𝑧  𝐼 then, (𝑎) all positive (𝑏) 1 positive and 2 negative.
Total number of ways = 180 + 3 𝐶2 × 180 = 720.

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Permutations and Combinations
Illustration 87:
Number of ways in which a pack of 52 playing cards be distributed equally among four players so that each
have the Ace, King, Queen and Jack of the same suit, is
36 ! . 4 ! 36 ! 52 ! . 4 ! 52 !
(A) (B) (C) (D)
( 9 !) ( 9 !) (13 !) (13 !)
4 4 4 4

Ans. (A)
Solution:
36 27 18 36!
Required number of ways 𝐶9 . 𝐶9 . 𝐶9 . 9 𝐶9 . 4 ! = × 4!
(9!)4
Illustration 88:
Find total number of positive integral solutions of 15 < 𝑥1 + 𝑥2 + 𝑥3 ≤ 20.
(A) 685 (B) 1140 (C) 455 (D) 1595
Ans. (A)
Solution:
𝑥1 + 𝑥2 + 𝑥3 = 20 – 𝑡
𝑡 = 0, 1, 2, 3, 4
4
Required value = 
t =0
19– t
C2 = 20
𝐶3 − 15
𝐶3 = 1140 − 455 = 685

Illustration 89:
Seven persons 𝑃1 , 𝑃2 , . . . . . . . . . , 𝑃7 initially seated at chairs 𝐶1 , 𝐶2 . . . . . . . . , 𝐶7 respectively. They all left their
chairs simultaneously for hand wash. Now in how many ways they can again take seats such that no one
sits on his own seat and 𝑃1 sits on 𝐶2 and 𝑃2 sits on 𝐶3 ?
(A) 52 (B) 53 (C) 54 (D) 55
Ans. (B)
Solution:
If 𝑃3 sits on 𝐶1
 1 1 1 1
4!  1– + – + 
 1! 2! 3! 4! 
= 4.3 − 4 + 1 = 9
If 𝑃3 does not sit on 𝐶1
 1 1 1 1 1
= 5!  1– + – + –  = 44
 1! 2! 3! 4! 5! 
total number of ways = 44 + 9 = 53
 P1 C1
P C → P
 2 2 1
 P3 C3 → P2

 P4 C4
P C
 5 5

 P6 C6

 P7 C7

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 JEE (Main + Advanced) : Mathematics
Illustration 90:
Given six line segments of length 2, 3, 4, 5, 6, 7 units, the number of triangles that can be formed by these
segments is
(A) 6 𝐶3 − 7 (B) 6 𝐶3 − 6 (C) 6 𝐶3 − 5 (D) 6 𝐶3 − 4
Ans. (A)
Solution:
First we select 3 length from the given 6 length so the no. of ways = 6 𝐶3
But these some pair i.e. (2, 3, 7), (2, 3, 6), (2, 3, 5) (2, 4, 6), (2, 4, 7), (2, 5, 7), (3, 4, 7) are not form a triangle
so that total no. of ways is 6 𝐶3 − 7 ways
Illustration 91:
In how many ways 𝐴, 𝐴, 𝐵, 𝐵, 𝐶, 𝐶, 𝐷, 𝐸, 𝐹, 𝐺 can be arranged around a circle. If no two identical letters are
together.
Solution:
𝐴, 𝐴, 𝐵, 𝐵, 𝐶, 𝐶, 𝐷, 𝐸, 𝐹, 𝐺 ⎯→ 10
10 − 1 𝐴
 ൬ ൰ −
2⋅ 2⋅ 2 𝐶
𝐵

𝐴 𝐵 𝐶
𝑛(𝐴) + 𝑛(𝐵) + 𝑛(𝐶) − 𝑛(𝐴 ∩ 𝐵) − 𝑛(𝐵 ∩ 𝐶) − (𝐶 ∩ 𝐴) + 𝑛(𝐴 ∩ 𝐵 ∩ 𝐶)
 9   9 − 1   8 –1  
  2·2·2  –   3–    3 + ( 7 –1 ) 
   2 · 2   2  
  
B C C
C A A
A B B

Illustration 92:
There are 𝑚 apples and 𝑛 oranges to be placed in a line such that the two extreme fruits being both oranges.
Let 𝑃 denotes the number of arrangements if the fruits of the same species are different and 𝑄 the
corresponding figure when the fruits of the same species are alike, then the ratio 𝑃/𝑄 has the value equal
to:
(A) 𝑛 𝑃2 . 𝑚 𝑃𝑚 . (𝑛 − 2) ! (B) 𝑚 𝑃2 . 𝑛 𝑃𝑛 . (𝑛 − 2) !
(C) 𝑛 𝑃2 . 𝑛 𝑃𝑛 . (𝑚 − 2) ! (D) none
Ans. (A)
Solution:
For 𝑷 → If same species are different
Total number of arrangements is 𝑛 𝑃2 . (𝑚 + 𝑛 − 2) !
( m + n − 2)!
For 𝑸 → If same species are alike then number of arrangement is
m ! .( n − 2)!
P
Hence = 𝑛 𝑃2 . 𝑚! . (𝑛 − 2)! = 𝑛
𝑃2 . 𝑚
𝑃𝑚 . (𝑛 − 2)!
Q

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Permutations and Combinations
Illustration 93:
The number of intersection points of diagonals of 2009 sides regular polygon, which lie inside the polygon.
(A) 2009 𝐶4 (B) 2009 𝐶2 (C) 2008 𝐶4 (D) 2008 𝐶2
Ans. (A)
Solution:
We know that in odd sides polygon no two or more then two diagonals are parallel, so if we take any 4
vertices, we get one point of intersection of diagonals.
Hence required no of points will be 2009 𝐶4 .
Illustration 94:
A rectangle with sides 2𝑚 – 1 and 2𝑛 – 1 is divided into squares of unit length by drawing parallel lines as
shown in the diagram, then the number of rectangles possible with odd side lengths is
2𝑛 − 1

2
1

1 2 2𝑚 − 1
(A) (𝑚 + 𝑛– 1)2 (B) 4𝑚+𝑛–1 (C) 𝑚2 𝑛2 (D) 𝑚(𝑚 + 1)𝑛(𝑛 + 1)
Ans. (C)
Solution:
2𝑛 − 1
2𝑛 − 2

3
2
1 2𝑚 − 2
0 2𝑚 − 1
0 1 2 3 2𝑚 − 3
No. of ways of choosing horizontal side of rectangle of one unit length = 2𝑚 − 1
No. of ways of choosing horizontal side of rectangle of 3 unit length = 2𝑚 − 3
 Total no. of ways of choosing horizontal side of rectangle of odd length
= (2𝑚– 1) + (2𝑚– 3)+. . . +1 = 𝑚2
Similarly no. of ways of choosing the vertical side of rectangle of odd length = 𝑛2.
 Total no. of ways of choosing the rectangle = 𝑛2 𝑚2

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 JEE (Main + Advanced) : Mathematics

EXERCISE - O
SINGLE CORRECT TYPE QUESTIONS
6

𝟏. The value of 50𝐶4 + ∑ 56−𝑟𝐶3 is

𝑟=1
56 56
(A) 𝐶4 (B) 𝐶3 (C) 55 𝐶3 (D) 55
𝐶4
MPC002
2. At an election, a voter may vote for any number of candidates, not greater than the number to be
elected. There are 10 candidates and 4 are to be elected. If a voter votes for at least one candidates,
then the number of ways in which he can vote is
(A) 385 (B) 1110 (C) 5040 (D) 6210
MPC003
3. The set 𝑆 = {1, 2, 3, . . . . . ,12} is to partitioned into three sets 𝐴, 𝐵, 𝐶 of equal size.
Thus, 𝐴  𝐵  𝐶 = 𝑆, 𝐴  𝐵 = 𝐵  𝐶 = 𝐶  𝐴 = , then number of ways to partition 𝑆 are-
12! 12! 12! 12!
(A) 4 (B) 3 (C) 4 (D) 3
3!(3!) (4!) (3!) 3!(4!)
MPC004
4. The number of integers greater than 6000 that can be formed, using the digits 3,5,6,7 and 8 without
repetition, is :
(A) 120 (B) 72 (C) 216 (D) 192
MPC005
5. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected
and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of
such arrangements is :
(A) At least 750 but less than 1000 (B) At least 1000
(C) Less than 500 (D) At least 500 but less than 750
MPC006
6. There are two urns. Urn 𝐴 has 3 distinct red balls and urn 𝐵 has 9 distinct blue balls. From each urn
two balls are taken out at random and then transferred to the other. The number of ways in which
this can be done is:
(A) 3 (B) 36 (C) 66 (D) 108
MPC007
7. If the letters of the word ′𝑆𝐴𝐶𝐻𝐼𝑁′ are arranged in all possible ways and these words are written
out as in dictionary, then the word ′𝑆𝐴𝐶𝐻𝐼𝑁′ appears at serial number
(A) 602 (B) 603 (C) 600 (D) 601
MPC001
8. The number of points, having both co-ordinates as integers, that lie in the interior of the triangle
with vertices (0, 0), (0, 41) and (41, 0) is :
(A) 820 (B) 780 (C) 901 (D) 861
MPC008

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Permutations and Combinations
9. There are 10 points in a plane, out of these 6 are collinear. If 𝑁 is the number of triangles formed
by joining these points, then:
(A) 𝑁 > 190 (B) 𝑁  100 (C) 100 < 𝑁  140 (D) 140 < 𝑁  190
MPC009
10. Let 𝑇𝑛 be the number of all possible triangles formed by joining vertices of an 𝑛 −sided regular
polygon. If 𝑇𝑛+1 − 𝑇𝑛 = 10, then the value of 𝑛 is :
(A) 7 (B) 5 (C) 10 (D) 8
MPC012
11. Assuming the balls to be identical except for difference in colours, the number of ways in which one
or more balls can be selected from 10 white, 9 green and 7 black balls is :
(A) 879 (B) 880 (C) 629 (D) 630
MPC010
12. Let 𝐴 and 𝐵 be two sets containing 2 elements and 4 elements respectively. The number of subsets
of 𝐴 × 𝐵 having 3 or more elements is
(A) 256 (B) 220 (C) 219 (D) 211
MPC011
13. A 5 digit number divisible by 3 is to be formed using the numerals 0,1,2,3,4 & 5 without repetition.
The total number of ways this can be done is –
(A) 3125 (B) 600 (C) 240 (D) 216
MPC013
14. Number of ways in which 9 different toys can be distributed among 4 children belonging to
different age groups in such a way that distribution among the 3 elder children is even and the
youngest one is to receive one toy more is-
(5!)2 9! 9!
(A) (B) 2 (C) 3 (D) none of these
8 3!(2!)
MPC014
15. The number of ways in which we can arrange 𝑛 ladies & 𝑛 gentlemen at a round table so that
2 ladies or 2 gentlemen may not sit next to one another is -
(A) (𝑛 − 1)! (𝑛 − 2)! (B) (𝑛)! (𝑛 − 1)!
(C) (𝑛 + 1)! (𝑛)! (D) none of these
MPC015
16. The number of way in which 10 identical apples can be distributed among 6 children so that each
child receives atleast one apple is -
(A) 126 (B) 252 (C) 378 (D) none of these
MPC016
17. The number of ways in which a mixed double tennis game can be arranged from amongst 9 married
couple if no husband & wife plays in the same game is -
(A) 756 (B) 3024 (C) 1512 (D) 6048
MPC017
18. Number of different words that can be formed using all the letters of the word “𝐷𝐸𝐸𝑃𝑀𝐴𝐿𝐴” if two
vowels are together and the other two are also together but separated from the first two is
(A) 960 (B) 1200 (C) 2160 (D) 1440
MPC018

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 JEE (Main + Advanced) : Mathematics
19. A road network as shown in the figure connect four cities. In how many ways can you start from
any city (say A) and come back to it without travelling on the same road more than once?
𝐴

𝐵 𝐶
(A) 8 (B) 12 (C) 9 (D) 16
MPC019
20. Sum of all the numbers that can be formed using all the digits 2, 3, 3, 4, 4, 4 is -
(A) 22222200 (B) 11111100 (C) 55555500 (D) 20333280
MPC020

MULTIPLE CORRECT TYPE QUESTIONS

21. Lines 𝑦 = 𝑥 + 𝑖 & 𝑦 = – 𝑥 + 𝑗 are drawn in 𝑥 – 𝑦 plane such that 𝑖  {1,2,3,4} & 𝑗{1,2,3,4,5,6}.
If 𝑚 represents the total number of squares formed by the lines and 𝑛 represents the total number
of triangles formed by the given lines & 𝑥-axis, then correct option/s is/are-
(A) 𝑚 + 𝑛 = 50 (B) 𝑚 − 𝑛 = 2 (C) 𝑚 + 𝑛 = 48 (D) 𝑚 − 𝑛 = 4
MPC052
22. The combinatorial coefficient 𝐶(𝑛, 𝑟) is equal to
(A) number of possible subsets of 𝑟 members from a set of 𝑛 distinct members.
(B) number of possible binary messages of length 𝑛 with exactly 𝑟 1's.
(C) number of non decreasing 2 − 𝐷 paths from the lattice point (0, 0) to (𝑟, 𝑛).
(D) number of ways of selecting 𝑟 things out of 𝑛 different things when a particular thing is always
included plus the number of ways of selecting ′𝑟′ things out of 𝑛, when a particular thing is
always excluded.
MPC053
23. There are 10 questions, each question is either True or False. Number of different sequences of
incorrect answers is also equal to
(A) Number of ways in which a normal coin tossed 10 times would fall in a definite order if both
Heads and Tails are present.
(B) Number of ways in which a multiple choice question containing 10 alternatives with one or
more than one correct alternatives, can be answered.
(C) Number of ways in which it is possible to draw a sum of money with 10 coins of different
denominations taken some or all at a time.
(D) Number of different selections of 10 indistinguishable things taken some or all at a time.
MPC054
24. Number of ways in which 3 numbers in A.P. can be selected from 1, 2, 3, . . . . . . 𝑛 is :
𝑛−1 2 𝑛(𝑛−2)
(A) ( 2 ) if 𝑛 is even (B) if 𝑛 is odd
4
(𝑛−1)2 𝑛(𝑛−2)
(C) if 𝑛 is odd (D) if 𝑛 is even
4 4
MPC055

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Permutations and Combinations
𝑛–1
25. The combinatorial coefficient 𝐶𝑝 denotes
(A) the number of ways in which 𝑛 things of which 𝑝 are alike and rest different can be arranged in
a circle.
(B) the number of ways in which 𝑝 different things can be selected out of 𝑛 different thing if a
particular thing is always excluded.
(C) number of ways in which 𝑛 alike balls can be distributed in 𝑝 different boxes so that no box
remains empty and each box can hold any number of balls.
(D) the number of ways in which (𝑛– 2) white balls and 𝑝 black balls can be arranged in a line if
black balls are separated, balls are all alike except for the colour.
MPC056
26. The maximum number of permutations of 2𝑛 letters in which there are only 𝑎′s & 𝑏's, taken all at
a time is given by:
2 6 10 4n − 6 4 n − 2
(A) 2𝑛 𝐶𝑛 (B) . . ........ .
1 2 3 n −1 n
n + 1 n + 2 n + 3 n + 4 2n − 1 2n 2n . 1.3.5....... (2n − 3)(2n − 1) 
(C) . . . ...... . (D)
1 2 3 4 n −1 n n!
MPC057

COMPREHENSION TYPE QUESTIONS

Paragraph for Question No. 27 to 29
Let 𝑝 be a prime number and 𝑛 be a positive integer, then exponent of 𝑝 in 𝑛! is denoted by 𝐸𝑝 (𝑛!)
and is given by
𝑛 𝑛 𝑛 𝑛
𝐸𝑝(𝑛!) = [ ] + [ 2 ] + [ 3 ] + ⋯ + [ 𝑘 ]
𝑝 𝑝 𝑝 𝑝
𝑘 𝑘+1
Where 𝑝 < 𝑛 < 𝑝
and [𝑥] denotes the integral part of 𝑥.
If we isolate the power of each prime contained in any number 𝑁, then 𝑁 can be written as
𝑁 = 2𝛼1 3𝛼2 5𝛼3 7𝛼4 . ..
where 𝛼𝑖 are whole numbers.

On the basis of above information, answer the following questions:

100
27. The exponent of 7 in 𝐶50 is -
(A) 0 (B) 1 (C) 2 (D) 3
MPC061
28. The number of zeros at the end of 108! is -
(A) 10 (B) 13 (C) 25 (D) 26
MPC062
29. The exponent of 12 in 100! is -
(A) 32 (B) 48 (C) 97 (D) none of these
MPC063

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 JEE (Main + Advanced) : Mathematics
MATRIX MATCH TYPE QUESTION
Answer Q.30, Q.31 and Q.32 by appropriately matching the information given in the three columns
of the following table
Column 1, 2 & 3 consists of different words, number of selection of 3 letters from the word and
number of arrangements of 3 letters from the word respectively.
Column-1 Column-2 Column-3
(I) DREAM (i) 70 (P) 60
(II) DEDICATION (ii) 10 (Q) 399
(III) POWERFUL (iii) 77 (R) 336
(IV) COMBINATION (iv) 56 (S) 378

30. Which of the following is a correct combination?

(A) (I) (i) (P) (B) (III) (iv) (R) (C) (II) (ii) (S) (D) (IV) (ii) (Q)
MPC058

31. Which of the following is correct combination?

(A) (III) (iv) (S) (B) (III) (i) (R) (C) (IV) (iii) (Q) (D) (I) (ii) (R)
MPC059

32. Which of the following is correct combination?

(A) (I) (ii) (S) (B) (II) (i) (Q) (C) (III) (iv) (R) (D) (IV) (iii) (R)
MPC060

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Permutations and Combinations

EXERCISE - S
1. An examination paper consists of 12 questions divided into parts 𝐴 & 𝐵. Part − 𝐴 contains 7
questions & Part-𝐵 contains 5 questions. If a candidate is required to attempt 8 questions selecting
atleast 3 from each part, then number of maximum ways can the candidate select the questions are
MPC024
2. Let 5 boys & 4 girls sit in a straight line. If the number of ways in which they can be seated if 2 girls
are together & the other 2 are also together but separated from the first 2 are 𝑁, then sum of digits
of 𝑁 is
MPC025
3. There are 2 women participating in a chess tournament. Every participant played 2 games with the
other participants. The number of games that the men played between themselves exceeded by 66
as compared to the number of games that the men played with the women. If number of
participants & the total number of games played in the tournament are 𝑀 and 𝑁 respectively, then
value of (𝑀 + 𝑁) is
MPC026
4. Find the ratio 𝐶𝑟 to 𝐶𝑟 when each of them has the greatest value possible.
20 25

MPC029
5. Let 𝑃𝑛 denotes the number of ways in which three people can be selected out of ‘𝑛’ people sitting in
a row, if no two of them are consecutive. If 𝑃𝑛+1 − 𝑃𝑛 = 15 then the value of ‘𝑛’ is
MPC021
6. The number of positive integral solutions of the equation 𝑥1 𝑥2 𝑥3 = 60 is
MPC022
7. Total number of even divisors of 2079000 which are divisible by 15 are
MPC023
8. If the sum of all numbers greater than 10000 formed by using the digits 0, 1, 2, 4, 5 & no digit being
repeated in any number, is 𝑁, then sum of digits of 𝑁 is
MPC027
9. Let the number of ways in which the letters of the word 'MUNMUN' can be arranged so that no two
alike letters are together are 𝑁, then value of 𝑁 is
MPC028
10. If number of integral solutions for the equation; 𝑥 + 𝑦 + 𝑧 + 𝑤 = 29 when
𝑁
𝑥 > 0, 𝑦 > 1, 𝑧 > 2 & 𝑤 ≥ 0 is equal to 𝑁, then find the value of
100?
MPC030
11. Six persons 𝐴, 𝐵, 𝐶, 𝐷, 𝐸 and 𝐹 are to be seated at a circular table. The number of ways this can be
done if 𝐴 must have either 𝐵 or 𝐶 on his right and 𝐵 must have either 𝐶 or 𝐷 on his right is -
MPC049
12. 5 Indian & 5 American couples meet at a party & shake hands. If no wife shakes hands with her own
husband & no Indian wife shakes hands with a male, then the number of hand shakes that takes
place in the party is –
MPC050
13. The number of ways of choosing a committee of 2 women & 3 men from 5 women & 6 men, if
Mr. 𝐴 refuses to serve on the committee if Mr. 𝐵 is a member & Mr. 𝐵 can only serve, if Miss 𝐶 is the
member of the committee, is -
MPC051

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 JEE (Main + Advanced) : Mathematics

EXERCISE - JEE (Main) PYQ

1. The number of four-digit numbers strictly greater than 4321 that can be formed using the digits
0,1,2,3,4,5 (repetition of digits is allowed) is : [JEE (Main) 2019]
(1) 288 (2) 306 (3) 360 (4) 310
MPC041
2. A committee of 11 members is to be formed from 8 males and 5 females. If 𝑚 is the number of ways
the committee is formed with at least 6 males and 𝑛 is the number of ways the committee is formed
with at least 3 females, then : [JEE (Main) 2019]
(1) 𝑚 = 𝑛 = 78 (2) 𝑛 = 𝑚 − 8 (3) 𝑚 + 𝑛 = 68 (4) 𝑚 = 𝑛 = 68
MPC042
3. Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of
one ball, the second row consists of two balls and so on. If 99 more identical balls are added to the
total number of balls used in forming the equilateral triangle, then all these balls can be arranged
in a square whose each side contains exactly 2 balls less than the number of balls each side of the
triangle contains. Then the number of balls used to form the equilateral triangle is :-
[JEE (Main) 2019]
(1) 190 (2) 262 (3) 225 (4) 157
MPC091
4. The number of natural numbers less than 7,000 which can be formed by using the digits 0,1,3,7,9
(repetition of digits allowed) is equal to : [JEE (Main) 2019]
(1) 250 (2) 374 (3) 372 (4) 375
MPC092
5. Let 𝑛 > 2 be an integer. Suppose that there are 𝑛 Metro stations in a city located along a circular
path. Each pair of stations is connected by a straight track only. Further, each pair of nearest
stations is connected by blue line, whereas all remaining pairs of stations are connected by red line.
If the number of red lines is 99 times the number of blue lines, then the value of 𝑛 is :
[JEE (Main) 2020]
(1) 199 (2) 101 (3) 201 (4) 200
MPC043
6. There are 3 sections in a question paper and each section contains 5 questions. A candidate has to
answer a total of 5 questions, choosing at least one question from each section. Then the number
of ways, in which the candidate can choose the questions, is : [JEE (Main) 2020]
(1) 1500 (2) 2255 (3) 3000 (4) 2250
MPC044
7. If the number of five digit numbers with distinct digits and 2 at the 10 place is 336 𝑘, then 𝑘 is
th

equal to : [JEE (Main) 2020]

(1) 8 (2) 6 (3) 4 (4) 7
MPC093
8. If the letters of the word 'MOTHER' be permuted and all the words so formed (with or without
meaning) be listed as in a dictionary, then the position of the word 'MOTHER' is ______.
[JEE (Main) 2020]
MPC094
9. If the digits are not allowed to repeat in any number formed by using the digits 0, 2, 4, 6, 8, then the
number of all numbers greater than 10,000 is equal to _______. [JEE (Main) 2021]
MPC095

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Permutations and Combinations
10. There are 5 students in class 10, 6 students in class 11 and 8 students in class 12. If the number of
ways, in which 10 students can be selected from them so as to include at least 2 students from each
class and at most 5 students from the total 11 students of class 10 and 11 is 100 𝑘, then 𝑘 is equal to.
[JEE (Main) 2021]
MPC096
11. The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit
"1" and they all are multiple of 11, is _______. [JEE (Main) 2021]
MPC097
12. The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is:
[JEE (Main) 2021]
(1) 26664 (2) 122664 (3) 122234 (4) 22264
MPC098
13. The letters of the word ‘MANKIND’ are written in all possible orders and arranged in serial order
as in an English dictionary. Then the serial number of the word ‘MANKIND’ is _______.
[JEE (Main) 2022]
MPC099
14. There are ten boys 𝐵1 , 𝐵2 , ...., 𝐵10 and five girls 𝐺1 , 𝐺2 , ...., 𝐺5 in a class. Then the number of ways of
forming a group consisting of three boys and three girls, if both 𝐵1 and 𝐵2 together should not be
the members of a group, is__________. [JEE (Main) 2022]
MPC100
15. Let 𝑆 be the set of all passwords which are six to eight characters long, where each character is
either an alphabet from {𝐴, 𝐵, 𝐶, 𝐷, 𝐸} or a number from {1, 2, 3, 4, 5} with the repetition of
characters allowed. If the number of passwords in 𝑆 whose at least one character is a number from
{1, 2, 3, 4, 5} is 56 , then  is equal to _____. [JEE (Main) 2022]
MPC101
16. The total number of four digit numbers such that each of the first three digits is divisible by the last
digit, is equal to______. [JEE (Main) 2022]
MPC102
17. The number of ways , in which 5 girls and 7 boys can be seated at a round table so that no two girls
sit together, is [JEE (Main) 2023]
(1) 126 (5!) (2) 7 (360) (4) 7 (720)
2 2 2
(3) 720
MPC103
18. A person forgets his 4-digit ATM pin code. But he remembers that in the code all the digits are
different, the greatest digit is 7 and the sum of the first two digits is equal to the sum of the last two
digits. Then the maximum number of trials necessary to obtain the correct code is____
[JEE (Main) 2023]
MPC104
19. The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1,3,5,7,9
without repetition, is [JEE (Main) 2023]
(1) 6 (2) 12 (3) 120 (4) 72
MPC105
20. If the number of words, with or without meaning, which can be made using all the letters of the
word MATHEMATICS in which 𝐶 and 𝑆 do not come together, is (6!)k, then 𝑘 is equal to
[JEE (Main) 2023]
(1) 1890 (2) 945 (3) 2835 (4) 5670
MPC106

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 JEE (Main + Advanced) : Mathematics

EXERCISE - JEE (Advanced) PYQ

1. A pack contains 𝑛 cards numbered from 1 to 𝑛. Two consecutive numbered cards are removed from
the pack and the sum of the numbers on the remaining cards is 1224. If the smaller to the numbers
on the removed cards is 𝑘, then 𝑘 – 20 = [JEE (Advanced) 2013]
MPC107
2. Let n1  n2  n3  n4  n5 be positive integers such that n1 + n2 + n3 + n4 + n5 = 20 . The number of such
distinct arrangements ( n1 , n2 , n3 , n4 , n5 ) is [JEE (Advanced) 2014]
MPC108
3. Let 𝑛 > 2 b an integer. Take 𝑛 distinct points on a circle and join each pair of points by a line segment.
Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the
number of red and blue line segments are equal, then the value of 𝑛 is [JEE (Advanced) 2014]
MPC109
4. Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so
that each envelope contains exactly one card and no card is placed in the envelope bearing the same
number and moreover the card numbered 1 in always placed in envelope numbered 2. Then the
number of ways it can be done is - [JEE (Advanced) 2014]
(A) 264 (B) 265 (C) 53 (D) 67
MPC110
5. Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all
the girls stand consecutively in the queue. Let 𝑚 be the number of ways in which 5 boys and 5 girls
can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then
m
the value of is [JEE (Advanced) 2015]
n
MPC111
6. A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club
including the selection of a captain (from among these 4 member) for the team. If the team has to
include at most one boy, then the number of ways of selecting the team is
[JEE (Advanced) 2016]
(A) 380 (B) 320 (C) 260 (D) 95
MPC064
7. Words of length 10 are formed using the letters 𝐴, 𝐵, 𝐶, 𝐷, 𝐸, 𝐹, 𝐺, 𝐻, 𝐼, 𝐽. Let 𝑥 be the number of such
words where no letter is repeated; and let 𝑦 be the number of such words where exactly one letter
𝑦
is repeated twice and no other letter is repeated. Then, = [JEE (Advanced) 2017]
9𝑥
MPC065
8. Let 𝑆 = {1, 2, 3,....., 9}. For 𝑘 = 1,2, . . . . . , 5, let 𝑁𝑘 be the number of subsets of 𝑆, each containing five
elements out of which exactly 𝑘 are odd. Then 𝑁1 + 𝑁2 + 𝑁3 + 𝑁4 + 𝑁5 =
[JEE (Advanced) 2017]
(A) 125 (B) 252 (C) 210 (D) 126
MPC066

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Permutations and Combinations
9. The number of 5 digit numbers which are divisible by 4, with digits from the set {1, 2, 3, 4, 5} and the
repetition of digits is allowed, is. [JEE (Advanced) 2018]
MPC067
10. Let 𝑋 be a set with exactly 5 elements and 𝑌 be a set with exactly 7 elements. If  is the number of
one-one functions from 𝑋 to 𝑌 and  is the number of onto functions from 𝑌 to 𝑋, then the value of
1
(𝛽 − 𝛼) is _____ . [JEE (Advanced) 2018]
5!
MPC068
11. In a high school, a committee has to be formed from a group of 6 boys 𝑀1 , 𝑀2 , 𝑀3 , 𝑀4 , 𝑀5 , 𝑀6 and
5 girls 𝐺1 , 𝐺2 , 𝐺3 , 𝐺4 , 𝐺5 .
(i) Let 𝛼1 be the total number of ways in which the committee can be formed such that the
committee has 5 members, having exactly 3 boys and 2 girls.
(ii) Let 𝛼2 be the total number of ways in which the committee can be formed such that the
committee has at least 2 members, and having an equal number of boys and girls.
(iii) Let 𝛼3 be the total number of ways in which the committee can be formed such that the
committee has 5 members, at least 2 of them being girls.
(iv) Let 𝛼4 be the total number of ways in which the committee can be formed such that the
committee has 4 members, having at least 2 girls and such that both 𝑀1 and 𝐺1 are NOT in the
committee together.
List - I List - II
P. The value of 𝛼1 is 1. 136
Q. The value of 𝛼2 is 2. 189
R. The value of 𝛼3 is 3. 192
S. The value of 𝛼4 is 4. 200
5. 381
6. 461
[JEE (Advanced) 2018]
The correct option is :
(A) P → 4; Q → 6, R → 2; S → 1
(B) P → 1; Q → 4; R → 2; S → 3
(C) P → 4; Q → 6, R → 5; S → 2
(D) P → 4; Q → 2; R → 3; S → 1
MPC069
12. Five person 𝐴, 𝐵, 𝐶, 𝐷 and 𝐸 are seated in a circular arrangement. If each of them is given a hat of
one of the three colours red, blue and green, then the number of ways of distributing the hats such
that the persons seated in adjacent seats get different coloured hats is [JEE (Advanced) 2019]
MPC070
13. An engineer is required to visit a factory for exactly four days during the first 15 days of every
month and it is mandatory that no two visits take place on consecutive days. Then the number of
all possible ways in which such visits to the factory can be made by the engineer during 1–15 June
2021 is _____. [JEE (Advanced) 2020]
MPC071

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 JEE (Main + Advanced) : Mathematics
14. In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in
such a way that each of these rooms contains at least one person and at most two persons. Then
the number of all possible ways in which this can be done is [JEE (Advanced) 2020]
MPC072
15. Let
𝑆1 = {(𝑖, 𝑗, 𝑘) : 𝑖, 𝑗, 𝑘  {1, 2,…,10}}
𝑆2 = {(𝐼, 𝑗) : 1  I < 𝑗 + 2  10, 𝐼, 𝑗  {1, 2,…,10}},
𝑆3 = {(𝑖, 𝑗, 𝑘, 𝑙) : 1  𝑖 < 𝑗 < 𝑘 < 𝑙, 𝑖, 𝑗, 𝑘, 𝑙  {1, 2,….,10}}.
and
𝑆4 = {(𝑖, 𝑗, 𝑘, 𝑙) : 𝑖, 𝑗, 𝑘 and 𝑙 are distinct elements in {1, 2,…,10}}.
If the total number of elements in the set 𝑆𝑟 is 𝑛𝑟 , 𝑟 = 1, 2, 3, 4, then which of the following
statements is (are) TRUE? [JEE (Advanced) 2021]
n
(A) 𝑛1 = 1000 (B) 𝑛2 = 44 (C) 𝑛3 = 220 (D) 4 = 420
12
MPC088
16. The number of 4-digit integers in the closed interval [2022, 4482] formed by using the digits
0, 2, 3, 4, 6, 7 is. [JEE (Advanced) 2022]
MPC089
17. Consider 4 boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are
distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each
box at least one red ball and one blue ball are chosen? [JEE (Advanced) 2022]
(A) 21816 (B) 85536 (C) 12096 (D) 156816
MPC090
18. Let 𝑋 be the set of all five digit numbers formed using 1,2,2,2,4,4,0. For example, 22240 is in 𝑋 while
02244 and 44422 are not in 𝑋. Suppose that each element of 𝑋 has an equal chance of being chosen.
Let 𝑝 be the conditional probability that an element chosen at random is a multiple of
20 given that it is a multiple of 5. Then the value of 38𝑝 is equal to [JEE (Advanced) 2023]
MPC112

[ 56 ] www.allen.in
Permutations and Combinations

JEE (Main) Practice Paper

This paper is for yourself practice and assessment the discussion of this paper is optional though you can
see PDF solutions or video solutions or solutions in hardcopy whichever is provided.
SECTION–A
• This section contains TWENTY questions.
• Each question has FOUR options (1), (2), (3) and (4). ONLY ONE of these four options is correct.
• For each question, darken the bubble corresponding to the correct option in the ORS.
• For each question, marks will be awarded in one of the following categories:
Full Marks : +4, if only the bubble corresponding to the correct option is darkened.
Zero Marks : 0, if none of the bubbles is darkened.
Negative Marks : –1 in all other cases.

1. Consider the five points comprising of the vertices of a square and the intersection point of its
diagonals. How many triangles can be formed using these points ?
(1) 4 (2) 6 (3) 8 (4) 10
MPC113
2. The kindergarten teacher has 25 kids in her class . She takes 5 of them at a time, to zoological garden
as often as she can, without taking the same 5 kids more than once. Then the number of visits, the
teacher makes to the garden exceeds that of a kid by :
(1) 25𝐶 5 − 24𝐶 5 (2) 24𝐶 5 (3) 24𝐶 4 (4) none
MPC114
3. Three vertices of a convex 𝑛 sided polygon are selected. If the number of triangles that can be
constructed such that none of the sides of the triangle is also the side of the polygon is 30, then the
polygon is a
(1) Heptagon (2) Octagon (3) Nonagon (4) Decagon
MPC115
4. If the letters of the word “VARUN” are written in all possible ways and then are arranged as in a
dictionary, then the rank of the word VARUN is :
(1) 98 (2) 99 (3) 100 (4) 101
MPC116
5. Number of 5 digit numbers which are divisible by 5 and each number containing the digit 5, digits
being all different is equal to 𝑘(4!), the value of 𝑘 is
(1) 84 (2) 168 (3) 188 (4) 208
MPC117
6. A rack has 5 different pairs of shoes. The number of ways in which 4 shoes can be chosen from it,
so that there will be no complete pair is :
(1) 1920 (2) 200 (3) 110 (4) 80
MPC118
7. Number of ways in which 8 people can be arranged in a line if 𝐴 and 𝐵 must be next each other and
𝐶 must be somewhere behind 𝐷, is equal to
(1) 10080 (2) 5040 (3) 5050 (4) 10100
MPC119

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 JEE (Main + Advanced) : Mathematics
8. Number of 5 digit numbers divisible by 25 that can be formed using only the digits 1, 2, 3, 4, 5 & 0
taken five at a time is
(1) 2 (2) 32 (3) 42 (4) 52
MPC120
9. In a unique hockey series between India & Pakistan, they decide to play on till a team wins
5 matches. The number of ways in which the series can be won by India, if no match ends in a draw
is :
(1) 126 (2) 252 (3) 225 (4) none
MPC121
10. Number of permutations of 1, 2, 3, 4, 5, 6, 7, 8 and 9 taken all at a time, such that the digit
1 appearing somewhere to the left of 2
3 appearing to the left of 4 and
5 somewhere to the left of 6, is
(e.g. 815723946 would be one such permutation)
(1) 9 · 7! (2) 8! (3) 5! · 4! (4) 8! · 4!
MPC122
11. A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only
4. If internal arrangement inside the car does not matter then the number of ways in which they
can travel, is
(1) 91 (2) 182 (3) 126 (4) 3920
MPC123
12. There are (𝑝 + 𝑞) different books on different topics in Mathematics. (𝑝  𝑞)
If 𝐿 = The number of ways in which these books are distributed between two students 𝑋 and 𝑌 such
that 𝑋 get 𝑝 books and 𝑌 gets 𝑞 books.
𝑀 = The number of ways in which these books are distributed between two students 𝑋 and 𝑌 such
that one of them gets 𝑝 books and another gets 𝑞 books.
𝑁 = The number of ways in which these books are divided into two groups of 𝑝 books and 𝑞 books
then,
(1) 𝐿 = 𝑀 = 𝑁 (2) 𝐿 = 2𝑀 = 2𝑁 (3) 2𝐿 = 𝑀 = 2𝑁 (4) 𝐿 = 𝑀 = 2𝑁
MPC124
13. The number of ways in which 8 distinguishable apples can be distributed among 3 boys such that
every boy should get atleast 1 apple & atmost 4 apples is 𝐾 · 7𝑃3 where 𝐾 has the value equal to
(1) 14 (2) 66 (3) 44 (4) 22
MPC125
14. There are six periods in each working day of a school. Number of ways in which 5 subjects can be
arranged if each subject is allotted at least one period and no period remains vacant is
(1) 210 (2) 1800 (3) 360 (4) 3600
MPC126
15. Number of positive integral solutions satisfying the equation (𝑥 1 + 𝑥 2 + 𝑥 3) (𝑦1 + 𝑦2) = 77, is
(1) 150 (2) 270 (3) 420 (4) 1024
MPC127

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Permutations and Combinations
16. If 𝑛 identical dice are rolled, then number of possible out comes are.
6n
(1) 6𝑛 (2) (3) ( n + 5)
c5 (4) None of these
n!
MPC128
17. Number of ways in which a pack of 52 playing cards be distributed equally among four players so
that each have the Ace, King, Queen and Jack of the same suit, is
36 ! . 4 ! 36 ! 52 ! . 4 ! 52 !
(1) (2) (3) (4)
( 9 !) ( 9 !) (13 !) (13 !)
4 4 4 4

MPC129
18. Seven person 𝑃1, 𝑃2, ........., 𝑃7 initially seated at chairs 𝐶 1, 𝐶 2 ........ , 𝐶 7 respectively. They all left their
chairs simultaneously for hand wash. Now in how many ways they can again take seats such that
no one sits on his own seat and 𝑃1 sits on 𝐶 2 and 𝑃2 sits on 𝐶 3 ?
(1) 52 (2) 53 (3) 54 (4) 55
MPC130
19. Given six line segments of length 2, 3, 4, 5, 6, 7 units, the number of triangles that can be formed by
these segments is
(1) 6𝐶 3 – 7 (2) 6𝐶 3 – 6 (3) 6𝐶 3 – 5 (4) 6𝐶 3 – 4
MPC131
m
20. Find the number of all rational number such that
n
m
(i) 0 < < 1, (ii) 𝑚 and 𝑛 are relatively prime (iii) 𝑚𝑛 = 25!
n
(1) 256 (2) 128 (3) 512 (4) None of these
MPC132

SECTION-B
 This section will have TEN questions. Candidate can choose to attempt any 5 question out of these
10 questions. In case if candidate attempts more than 5 questions, first 5 attempted questions will
be considered for marking.
 The answer to each question is a NUMERICAL VALUE.
 For each question, enter the correct numerical value (Answer should be rounded off to the nearest
integer).
 Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4, if only correct answer is given.
Zero Marks : 0, if no answer is given.
Negative Marks : -1 for incorrect answer

1. Number of five digits numbers divisible by 3 that can be formed using the digits 0, 1, 2, 3, 4, 7 and
N
8 if, each digit is to be used atmost one is 𝑁 then   is equal to
8
MPC133

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 JEE (Main + Advanced) : Mathematics
2. The sides 𝐴𝐵, 𝐵𝐶 & 𝐶𝐴 of a triangle 𝐴𝐵𝐶 have 3, 4 & 5 interior points respectively on them. If the
k
number of triangles that can be constructed using these interior points as vertices is 𝑘, then  
5
is equal to
MPC134
3. Seven different coins are to be divided amongst three persons. If no two of the persons receive the
same number of coins but each receives atleast one coin & none is left over, then the number of
 k 
ways in which the division may be made is 𝑘, then   is equal to.
 10 
MPC135
4. The number of integers which lie between 1 and 106 and which have the sum of the digits equal to
12 is 𝑁 then (𝑁–6000) is equal to
MPC136
5. In a hockey series between team 𝑋 and 𝑌, they decide to play till a team wins ‘10’ match. Then the
20
Cm
number of ways in which team 𝑋 wins is then m is equal to
2
MPC137
6. The number of permutations which can be formed out of the letters of the word "SERIES" taking
three letters together is
MPC138
7. A box contains 6 balls which may be all of different colours or three each of two colours or two each
of three different colours. The number of ways of selecting 3 balls from the box (if ball of same
colour are identical) is
MPC139
8. The number of ways selecting 8 books from a library which has 10 books each of Mathematics,
Physics, Chemistry and English, if books of the same subject are alike, is (𝑁2 – 4) then N is equal to
MPC140
9. Let 𝑃𝑛 denotes the number of ways in which three people can be selected out of '𝑛' people sitting
𝑛
in a row, if no two of them are consecutive. If, 𝑃𝑛+1 − 𝑃𝑛 = 15 then the value of 𝐶2 is :
MPC160
2002
10. Number of cyphers at the end of C1001 is
MPC141

[ 60 ] www.allen.in
Permutations and Combinations

JEE (Advanced) Practice Paper

This paper is for yourself practice and assessment the discussion of this paper is optional though you can
see PDF solutions or video solutions or solutions in hardcopy whichever is provided.
SECTION-I
• This section contains SIX questions.
• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.
• For each question, darken the bubble corresponding to the correct option in the ORS.
• For each question, marks will be awarded in one of the following categories:
Full Marks : +3 If only the bubble corresponding to the correct option is darkened.
Zero Marks : 0 If none of the bubbles is darkened.
Negative Marks : –1 In all other cases

1. A bouquet from 11 different flowers is to be made so that it contains not less than three flowers.
Then then number of different ways of selecting flowers to form the bouquet.
(A) 1972 (B) 1952 (C) 1981 (D) 1947
MPC142
2. If  = x1 x2 x3 and  = y1 y2 y3 be two three digit numbers, then the number of pairs of  and  that

can be formed so that  can be subtracted from  without borrowing.

(A) 55 . (45)2 (B) 45 . (55)2 (C) 36 . (45)2 (D) 553
MPC143
3. '𝑛' digits positive integers formed such that each digit is 1, 2, or 3. How many of these contain all
three of the digits 1, 2 and 3 atleast once ?
(A) 3(𝑛 –1) (B) 3n − 2.2n + 3 (C) 3n − 3.2n − 3 (D) 3n − 3.2n + 3
MPC144
4. 𝑋 = {1, 2, 3, 4, ...... 2017} and 𝐴  𝑋 ; 𝐵  𝑋 ; 𝐴  𝐵  𝑋 here 𝑃  𝑄 denotes that 𝑃 is subset of
Q( P  Q ) . Then number of ways of selecting unordered pair of sets 𝐴 and 𝐵 such that 𝐴  𝐵  𝑋.

(42017 − 32017 ) + (22017 − 1) (42017 − 32017 )

(A) (B)
2 2
42017 − 32017 + 22017
(C) (D) None of these
2
MPC145
5. The number of ways in which 15 identical apples & 10 identical oranges can be distributed among
three persons, each receiving none, one or more is:
(A) 5670 (B) 7200 (C) 8976 (D) 7296
MPC146

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 JEE (Main + Advanced) : Mathematics
6. Two variants of a test paper are distributed among 12 students. Number of ways of seating of the
students in two rows so that the students sitting side by side do not have identical papers & those
sitting in the same column have the same paper is:
12! (12)!
(A) (B) (C) (6!)2. 2 (D) 12! × 2
6! 6! 25 . 6!
MPC147

SECTION-II
• This section contains SIX questions.
• Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these four
option(s) is (are) correct option(s).
• For each question, choose the correct option(s) to answer the question.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If only (all) the correct option(s) is (are) chosen.
Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.
Partial Marks : +2 If three or more options are correct but ONLY two options are chosen,
both of which are correct options.
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen
and it is a correct option.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).
Negative Marks : –2 In all other cases.
For Example : If first, third and fourth are the ONLY three correct options for a question with second
option being an incorrect option; selecting only all the three correct options will result in +4 marks.
Selecting only two of the three correct options (e.g. the first and fourth options), without selecting
any incorrect option (second option in this case), will result in +2 marks. Selecting only one of the
three correct options (either first or third or fourth option), without selecting any incorrect option
(second option in this case), will result in +1 marks. Selecting any incorrect option(s) (second option
in this case), with or without selection of any correct option(s) will result in –2 marks.
7. The number of ways in which 10 students can be divided into three teams, one containing 4 and
others 3 each, is
10! 10! 1
(A) (B) 2100 (C) 10𝐶 4 . 5𝐶 3 (D) .
4!3!3! 6!3!3! 2
MPC148
8. If letters of the word 'SAMANVAYA' are arranged around a circle then number of permutations is
n!
, then possible value of x + y − n is -
2 2

x! y!
(A) 13 (B) 12 (C) 8 (D) 9
MPC149
9. The number of five digit numbers that can be formed using all the digits 0, 1, 3, 6, 8 which are -
(A) divisible by 4 is 30
(B) greater than 30,000 and divisible by 11 is 12
(C) smaller than 60,000 when digit 8 always appears at ten's place is 6
(D) between 30,000 and 60,000 and divisible by 6 is 18.
MPC150

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Permutations and Combinations
10. All the 7 digit numbers containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once and not divisible
by 5 are arranged in the increasing order. Then -
(A) 1800𝑡ℎ number in the list is 3124567 (B) 1897𝑡ℎ number in the list is 4213567
(C) 1994𝑡ℎ number in the list is 4312567 (D) 2001𝑡ℎ number in the list is 4315726
MPC151
11. If 𝑎, 𝑏, 𝑐, 𝑑 are prime numbers such that a 2 − b2 = c & 𝑑 = 𝑏 + 𝑐, then sum of all the 4 digit numbers
formed by using the digits a, b, c, d without repetition is -
(A) 112233 (B) 113322
(C) an even number divisible by 11 (D) an odd number divisible by 11
MPC152
12. If 31! 33! 35! is divisible by 2m ( m  N ) , then value of 𝑚 can be :-
(A) 69 (B) 79 (C) 89 (D) 99
MPC153

SECTION–III
 This section contains THREE paragraphs.
• Based on each paragraph, there are TWO questions.
• Each question has FOUR options (A), (B), (C) and (D) ONLY ONE of these four options is correct.
• For each question, darken the bubble corresponding to the correct option in the ORS.
• For each question, marks will be awarded in one of the following categories :
Full Marks : +3 if only the bubble corresponding to the correct answer is darkened.
Zero Marks : 0 in all other cases.

Comprehension # 1 (Q. No. 13 - 14)

16 players P1 , P2 , P3 ,....... P16 take part in a tennis tournament. Lower suffix player is better than any
higher suffix player. These players are to be divided into 4 groups each comprising of 4 players and
the best from each group is selected for semi-finals.
13. Number of ways in which they can be divided into 4 equal groups if the players P1 , P2 , P3 and P4
are in different groups, is :
(11)! (11)! (11)! (11)!
(A) (B) (C) (D)
36 72 108 216
MPC154
14. Number of ways in which these 16 players can be divided into four equal groups, such that when
12!
the best player is selected from each group, 𝑃6 is one among them, is (𝑘) . The value of 𝑘 is :
(4!)3
(A) 36 (B) 24 (C) 18 (D) 20
MPC155

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 JEE (Main + Advanced) : Mathematics
Comprehension # 2 (Q. No. 15 - 16)
Consider the number 𝑁 = 2910600. On the basis of above information, answer the following
questions :
15. Total number of divisors of 𝑁, which are divisible by 15 but not by 36 are -
(A) 92 (B) 94 (C) 96 (D) 98
MPC156
16. Total number of ways, in which the given number can be split into two factors such that their
highest common factor is a prime number is equal to -
(A) 16 (B) 32 (C) 48 (D) 64
MPC157

Comprehension # 3 (Q. No. 17 - 18)

2 American men; 2 British men; 2 Chinese men and one each of Dutch, Egyptial, French and German
persons are to be seated for a round table conference.
17. If the number of ways in which they can be seated if exactly two pairs of persons of same nationality
p
are together is p(6!), then find .
100
2 3 4 6
(A) (B) (C) (D)
5 5 5 5
MPC158
18. If the number of ways in which only American pair is adjacent is equal to 𝑞(6!), then find.
16 18 21 23
(A) (B) (C) (D)
25 25 25 25
MPC159

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Permutations and Combinations

ANSWER KEY
EXERCISE - O

Que. 1 2 3 4 5 6 7 8 9 10
Ans. A A B D B D D B B B
Que. 11 12 13 14 15 16 17 18 19 20
Ans. A C D C B A C D B A
Que. 21 22 23 24 25 26 27 28 29 30
Ans. A,B A,B,D B,C C,D B,D A,B,C,D A C B B
Que. 31 32
Ans. C C

EXERCISE - S
1. 420 2. 9 3. 169 4. 420 5. 8
6. 54 7. 108 8. 36 9. 30 10. 26
11. 18 12. 135 13. 124

EXERCISE - JEE (Main) PYQ

Que. 1 2 3 4 5 6 7 8 9 10
Ans. 4 1 1 2 3 4 1 309 96 238
Que. 11 12 13 14 15 16 17 18 19 20
Ans. 7744 1 1492 1120 7073 1086 1 72 4 4

EXERCISE - JEE (Advanced) PYQ

Que. 1 2 3 4 5 6 7 8 9 10
Ans. 5 7 5 C 5 A 5 D 625 119
Que. 11 12 13 14 15 16 17 18
Ans. C 30 495 1080 A,B,D 569.00 A 31

JEE (Main) Practice Paper

Q. 1 2 3 4 5 6 7 8 9 10
A. 3 2 3 3 2 4 2 3 1 1
Section-A
Q. 11 12 13 14 15 16 17 18 19 20
A. 3 3 4 2 3 3 1 2 1 1
Q. 1 2 3 4 5 6 7 8 9 10
Section-B
A. 93 41 63 62 10 42 31 13 28 1

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 JEE (Main + Advanced) : Mathematics
JEE (Advanced) Practice Paper

Q. 1 2 3 4 5 6
Section-I
A. C B D A C D
Q. 7 8 9 10 11 12
Section-II
A. B,C C,D A,B,D B,D B,C A,B,C
Q. 13 14 15 16 17 18
Section-III
A. C D C C B A

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Permutations and Combinations

Permutations and Combinations

SOLUTIONS
EXERCISE - O
1. Ans. (A)
50
𝐶4 + 55 𝐶3 + 54 𝐶3 + 53 𝐶3 + 52 𝐶3 + 51 𝐶3 + 50 𝐶3
= 51 𝐶4 + 51 𝐶3 + 52 𝐶3 + 53 𝐶3 + 54 𝐶3 + 55 𝐶3 = 56 𝐶4
2. Ans. (A)
10
𝐶1 + 10 𝐶2 + 10 𝐶3 + 10 𝐶4 = 385
3. Ans. (B)
12! 12!
Number of ways = 3 3! =
(4!) .3! (4!)3
4. Ans. (D)
Number of 4 digit numbers greater than 6000 is 3 × 4 × 3 × 2 = 72
Number of 5 digit numbers greater than 6000 is 5! = 120
So total number of numbers = 72 + 120 = 192.
5. Ans. (B)
The no. of ways to select 4 novels & 1 dictionary from 6 different novels & 3 different dictionary
are 6 𝐶4 × 3 𝐶1
and to arrange these things in shelf so that dictionary is always in middle _ _ 𝐷 _ _ are 4!
Required No. of ways 6 𝐶4 × 3 𝐶1 × 4! = 1080
6. Ans. (D)
Urn 𝐴 → 3 Red balls
Urn 𝐵 → 9 Blue balls
So the number of ways = selection of 2 balls from urn 𝐴 & 𝐵 each.
= 3 𝐶2 . 9 𝐶2 = 108
7. Ans. (D)
𝐴 → 5!; 𝐶 → 5!; 𝐻 → 5!; 𝐼 → 5!; 𝑁 → 5!
SACHIN → 1
Rank = 5 × 5! + 1 = 601
8. Ans. (B)
Let 𝑃 be (𝑥, 𝑦)
𝑥1 (0, 41)
𝑦1
 𝑥 + 𝑦 < 41
𝑥 + 𝑦  40
𝑥 + 𝑦 + 𝑡 = 40, 𝑡  0 (0, 0) (41, 0)
(𝑥 − 1) + (𝑦 − 1) + 𝑡 = 38
 No. of points = 38+3–1 𝐶3–1 = 40 𝐶2 = 780
9. Ans. (B)
10×9×8 6×5×4
𝑁 = 10 𝐶3 − 6 𝐶3 = 3×2×1 − 3×2×1 = 120 − 20 = 100
𝑁  100

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 JEE (Main + Advanced) : Mathematics
10. Ans. (B)
𝑛 𝑛
𝑇𝑛 = 𝐶3 ⇒ 𝑛+1 𝐶3 − 𝐶3 = 10
(𝑛 + 1)𝑛(𝑛 − 1) − 𝑛(𝑛 − 1)(𝑛 − 2) = 60
𝑛(𝑛 − 1) = 20  𝑛 = 5
11. Ans. (A)
𝑊 10 , 𝐺 9 , 𝐵7
Selection of one or more balls
= (10 + 1)(9 + 1)(7 + 1) − 1 = 11 × 10 × 8 − 1 = 879
12. Ans. (C)
(𝐴, 𝐵)
 
2 × 4 = 8
8
𝐶3 + 8 𝐶4 +. . . . . + 8 𝐶8
= 28 − 8 𝐶0 − 8 𝐶1 − 8 𝐶2 = 256 − 37 = 219
13. Ans. (D)
Case – I, digits used are 1,2,3,4,5 = 5! = 120
Case – II, when digits used are 0,1,4,5,2 = 5! – 4! = 96
 Total = 120 + 96 = 216
14. Ans. (C)
9 different toys

2 3
2 2
9!
= × 3!
(2!)3 .3!.3!
15. Ans. (B)
First we arrange n ladies by (𝑛 − 1)! ways.
Thus we get 𝑛 gaps between them. Now arrange 𝑛 gentelmen among them
So total (𝑛 − 1)! × 𝑛! ways
16. Ans. (A)
𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 + 𝑥6 = 10
Giving one-one apples to each
𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 + 𝑥6 = 4
Here, 𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 , 𝑥5 , 𝑥6  0
So total 9 𝐶5 = 126 ways
17. Ans. (C)
9C2 × 7C2 × 2 = 1512
Team formation
Selection of female
Selection of male
18. Ans. (D)
Case-I : 𝐷, 𝑃, 𝑀, 𝐿. 𝐸𝐸 , 𝐴𝐴 → 4! × 5𝐶2 × 2!
Case-II : 𝐷, 𝑃, 𝑀, 𝐿, 𝐴𝐸 𝐴𝐸 → 4! × 5 𝐶2 × 2! × 2! × 1
Total = 1440

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Permutations and Combinations
19. Ans. (B)
Paths are shown as :
A

D–B–A
A–C A
B
D–A
4 paths
Similarly if we start from A towards D we get another 4 paths.
A

C–B–A
A–D
B–A
C–A
Similarly if we start from A towards B
Again 4 paths
Total different paths = 4 × 3 = 12
II Method → 3 𝐶1 × 2 𝐶1 × 2 𝐶1 = 12
20. Ans. (A)
6!
Number of numbers formed by 2,3,3,4,4, 4 are = 3!2! = 60

In each column 2,3,4 occurs 10, 20 & 30 times respectively.

So sum of each column is 2 × 10 + 3 × 20
4 × 30 = 200
21. Ans. (A,B)

𝑚 = 15 + 8 + 3 = 26
𝑛 = 4 𝐶1 . 6 𝐶4 = 24
𝑚 + 𝑛 = 50
}
𝑚−𝑛 =2

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 JEE (Main + Advanced) : Mathematics
22. Ans. (A, B, D)
In (C) number = 𝑛+𝑟 𝐶𝑟 ;
(D) 𝑛 𝐶𝑟 = 𝑛–1 𝐶𝑟–1 + 𝑛–1 𝐶𝑟
23. Ans. (B, C)
210 − 1; (𝐴)210 − 2; (𝐵)210 − 1; (𝐶)210 − 1; (𝐷)10
24. Ans. (C, D)
𝑛 = 2𝑚, arrange the numbers into disjoint sets
1, 3, 5, .................... (2𝑚 − 1) 𝑚 number
2, 4, 6, ..................... 2𝑚 𝑚 numbers
No. of 𝐴𝑃′𝑠 = 𝑚 𝐶2 + 𝑚 𝐶2
25. Ans. (B, D)
(𝑛−1)!
(A)
𝑝!
𝑛–1
(B) 𝐶𝑝
(C) 𝐵1 + 𝐵2 +. . . +𝐵𝑃 = 𝑛
Put 1 ball in each box.
 𝐵1′ + 𝐵2′ +. . . . + 𝐵𝑝′ = 𝑛 – 𝑝
using beggars method
𝑛–𝑝+(𝑝–1) 𝑛–1
number of ways = 𝐶𝑝–1 = 𝐶𝑝–1
(D) balls of same colour are identical
 arrange white balls → no of ways = 1
gaps in white balls = (𝑛– 1)
Now 𝑝 black balls put in gaps
(𝑛–1)
 No of ways (D) = 𝐶𝑝
26. Ans. (A,B,C,D)
Say there are (𝑥)𝑎′s, (2𝑛 − 𝑥)𝑏's
(2𝑛)!
so no permutations = 2𝑛𝐶𝑥
𝑥!(2𝑛−𝑥)!
= 2𝑛𝐶𝑥 is max if 𝑥 = 𝑛
so no of max permutation = 2𝑛𝐶𝑛
2𝑛 (2𝑛)! 1.2.3.4...2𝑛
𝐶𝑛 = =
𝑛!𝑛! 𝑛!𝑛!
[1.3.5. . . (2𝑛 − 1)][2.4.6. . .2𝑛] [1.3.5. . . . (2𝑛 − 1)]2𝑛 [1.2.3. . . 𝑛]
= =
𝑛! 𝑛! 𝑛! 𝑛!
2𝑛
1.2.3. . . 𝑛(𝑛 + 1)(𝑛 + 2). . . (2𝑛)
𝐶𝑛 =
𝑛! 𝑛!
2𝑛
𝑛 + 1 𝑛 + 2 2𝑛
𝐶𝑛 = . …
1 2 𝑛
2𝑛
(1.2.3. . .2𝑛) [1.3.5. . . . (2𝑛 − 3)(2𝑛 − 1)][1.2.3. . . 𝑛]
𝐶𝑛 = =
𝑛! 𝑛! 𝑛! 𝑛!
2 𝑛 [1.3.5.
. . . (2𝑛 − 3)(2𝑛 − 1)][1.2.3. . . 𝑛] [(2.1)(2.3)(2.5). . . (2(2𝑛 − 3)(2(2𝑛 − 1)))]
= =
𝑛! 𝑛! 𝑛!
2𝑛
2 6 10 4𝑛 − 6 4𝑛 − 2
𝐶𝑛 = . . . . . . .
1 2 3 𝑛−1 𝑛

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Permutations and Combinations
27. Ans. (A)
Exponent of 7 in 100!
100 14
[ ] + [ ] = 14 + 2 = 16
7 7
exponent of 7 in 50!
50 7
[ ]+[ ] = 8
7 7
100! 716
Exponent of 7 in 100 𝐶50 = = 8 8 = 7°
50! 50! 7 7
 exponent of 7 will be 0.
28. Ans. (C)
Product of 5's & 2's constitute 0's at the end of a
number  No. of 0's in 108!
= exponent of 5 in 108!
(Note that exponent of 2 will be more than
exponent of 5 in 108 !)
108 21
⇒[ ] + [ ] = 21 + 4 = 25
5 5
29. Ans. (B)
As 12 = 22 . 3, here we have to calculate exponent of 2 and exponent of 3 in 100!
exponent of 2
100 50 25 12 6 3
= [ ] + [ ] + [ ] + [ ] + [ ] + [ ] = 97
2 2 2 2 2 2
100 33 11 3
exponent of 3 = [ ] + [ ] + [ ] + [ ] = 48
3 3 3 3
Now, 12 = 2 × 2 × 3
we require two 2's & one 3
 exponent of 3 will give us the exponent of 12 in 100! i.e. 48
30. Ans. (B)
31. Ans. (C)
32. Ans. (C)
Solution for Q.30 to Q.32
Selection Arrangement Correct combinations
5 5
Dream 𝐶3 = 10 𝐶3 × 3! = 60 (I) (ii) (P)
3!
Dedication 2𝐶1 7𝐶1 + 8𝐶3 = 70 2𝐶1 7𝐶1 × +8 𝐶3 × 3! = 378 (II) (i) (S)
2!
Powerful 8𝐶3 = 56 8 𝐶3 × 3! = 336 (III) (iv) (R)
3!
Combination 3𝐶1 7𝐶1 + 8𝐶3 = 77 3𝐶1 7𝐶1 × +8 𝐶3 × 3! = 399 (IV) (iii) (Q)
2!

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EXERCISE - S
1. Ans. (420)
A B
5 3
4 4
3 5
= 𝐶5 × 𝐶3 + 7 𝐶4 × 5 𝐶4 + 7 𝐶3 × 5 𝐶5 = 420
7 5

2. Ans. (9)
Step 1𝑠𝑡 : Arrange 5 boys in 5! ways
Step 2𝑛𝑑 ∶ Select 2 gaps from 6 gaps for 4 girls (2girls for each gap) in 6 𝐶2 ways.
Step 3𝑟𝑑 : Select 2 girls to sit in one of the gaps and other 2 in remaining selected gaps
= 4 𝐶2 ways
Step 4 : Arrange 1𝑠𝑡 , 2 girls in 2! and other 2 in 2! Ways
Hence, total ways (𝑁) = 5! × 6 𝐶2 × 4 𝐶2 × 2 × 2 = 43200
Sum of digits of 𝑁 = 4 + 3 + 2 + 0 + 0 = 9
3. Ans. (169)
Given: 𝑛 𝐶2 × 2 = 𝑛 𝐶1 × 2 𝐶1 × 2 + 66
where 𝑛 = number of men
𝑛
𝐶2 − 2.𝑛 𝐶1 = 33
𝑛 = 11
So total no. of participants (𝑀) = 11 + 2 = 13
Total no. of games played (𝑁)
= 11 𝐶2 × 2 + 2 𝐶1 × 11 𝐶1 × 2 + 2 = 156
4. Ans. (420)
20 20
𝐶𝑟 is maximum when 𝑟 = 2 = 10
25 25−1 25+1
𝐶𝑟 is maximum when 𝑟 = or 2
2
𝑟 = 12 or 13
20𝐶
10
So 25 = 420
𝐶12
5. Ans. (8)
𝑃𝑛 = 𝑛–3+1 𝐶3 = 𝑛–2 𝐶3
Now 𝑃𝑛+1 − 𝑃𝑛 = 15
(𝑛–1) (𝑛–2) (𝑛−1)! (𝑛−2)!
 𝐶3 − 𝐶3 = 15  (𝑛−4)! − (𝑛−5)! = 90
3
 (𝑛−4) ⋅ (𝑛 − 2)(𝑛 − 3)(𝑛 − 4) = 90
 (𝑛 − 2)(𝑛 − 3) = 30  𝑛2 − 5𝑛 − 24 = 0
𝑛=8
6. Ans. (54)
𝑥1 𝑥2 𝑥3 = 60
𝑥1 . 𝑥2 . 𝑥3 = 22 × 3 × 5
𝑥1 = 2𝑎1 ⋅ 3𝑎2 ⋅ 5𝑎3
𝑥2 = 2𝑏1 ⋅ 3𝑏2 ⋅ 5𝑏3

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Permutations and Combinations
𝑥3 = 2𝑐1 ⋅ 3𝑐2 ⋅ 5𝑐3
𝑎1 + 𝑏1 + 𝑐1 = 2 → 4 𝐶2 = 6
𝑎2 + 𝑏2 + 𝑐2 = 1 → 3 𝐶2 = 3
𝑎3 + 𝑏3 + 𝑐3 = 1 → 3 𝐶2 = 3
Total solutions = 6 × 3 × 3 = 54
7. Ans. (108)
𝑁 = 2079000
𝑁 = 23 × 33 × 53 × 7 × 11
Total no. of even divisiors which are divisible by 15.
= 3 × 3 × 3 × 2 × 2 = 108
8. Ans. (36)
5 different digits can be arranged in 5! ways
so each digit will appear every place = 24 times
Sum of all digits at unit place = 24(0 + 1 + 2 + 4 + 5) = 288
Sum of all digits = 288 × 11111 = 3199968
When 0 is at first place 4 digit number will be formed
Each number will appear 6 times at every place.
Sum of all 4 digit number at unit place
= 6(1 + 2 + 4 + 5) = 72
Hence sum of all four digit number
= 72 × 1111 = 79992
So required sum = 3199968 − 79992 = 3119976
 𝑁 = 36
9. Ans. (30)
𝑀′𝑠 = 2, 𝑈′𝑠 = 2, 𝑁′𝑠 = 2
6!
Total = = 90
2! ⋅ 2! ⋅ 2!
Let 𝑛(𝐸1 ) = When one alike letter is together.
𝑛(𝐸2 ) = When two alike letters are together.
𝑛(𝐸3 ) = When three alike letters are together.
𝑛(𝐸3 ) = 𝑀𝑀 𝑈𝑈 𝑁𝑁 = 3! = 6
 4! 
𝑛(𝐸2 ) = 3[𝑛(𝐴  𝐵) – 𝑛(𝐴  𝐵  𝐶)] = 3  − 6  = 18
 2! 
𝑛(𝐸1 ) = 3[𝑛(𝐴) − {𝑛(𝐴  𝐵) + 𝑛(𝐵  𝐶)} + 𝑛(𝐴  𝐵  𝐶)]
 5! 
= 3 − (12.2 − 6 )  = 330 − 18 = 36
 2!2! 
Hence required number of ways
= 90 − [6 + 18 + 36] = 30
10. Ans. (26)
𝑥 + 𝑦 + 𝑧 + 𝑤 = 29
when 𝑥 > 0, 𝑦 > 1, 𝑧 > 2, 𝑤 ≥ 0
𝑥  1, 𝑦  2, 𝑧  3, 𝑤  0
Now put
𝑥 = 1 + 𝑡, 𝑦 = 2 + 𝑡2 , 𝑧 = 3 + 𝑡3 , 𝑤 = 0 + 𝑡4

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where 𝑡1 , 𝑡2 , 𝑡3 , 𝑡4  0
𝑡1 + 𝑡2 + 𝑡3 + 𝑡4 = 23
23+4–1 26
So no. of integral solutions = 𝐶4–1 = 𝐶3 = 2600 = 𝑁
𝑁
= 26
100
11. Ans. (18)
A A
B C
1
1 1 1 1

2 2
3 C or D

Case-I: 2 × 3 × 2 = 12 Case-II: 3 𝐶1 × 2 × 1 = 6
Total = 18
12. Ans. (135)
Method-I
Total number of handshakes possible 20 𝐶2
undesirable handshakes : 10 𝐶2 + 10 𝐶1 25 ways
I H AH
Hence, Desired ways = 20 𝐶2 − ( 10 𝐶2 + 10 𝐶1 )
Method-II
𝐼𝐻 → Indian husband, 𝐼𝑊 → Indian wife
5 × 4 = 20 ways
𝐴𝐻 → American husband, 𝐴𝑊 → American wife
Case 1 : Hand shaking occurring between same
nationals and same genders (𝑀/𝐹)
IW AW
𝐼𝐻 − 𝐼𝐻 → 5 𝐶2 = 10 ways 5 × 5 = 25 ways
Similarly for 𝐼𝑊 − 𝐼𝑊 , 𝐴𝑊 − 𝐴𝑊 , 𝐴𝐻 − 𝐴𝐻
Total ways 10 × 4 = 40.
Case 2 : All other possible hand shakes
Hence total number of handshakes
= (25 × 3 + 40) + (20) = 135
13. Ans. (124)
Case 1 : Mr. 𝐵 & Miss 𝐶 are in committee and Mr. 𝐴 is excluded
 4 𝐶2 × 4 𝐶1 = 24 ways
(Men) (Women)
Case 2 : Mr. 𝐵 not there ∶ 5 𝐶3 × 5 𝐶2 = 100 ways
(Men) (Women)
Total ways = 24 + 100 = 124

EXERCISE - JEE (Main) PYQ

1. Ans. (4)
(1) The number of four-digit numbers Starting with 5 is equal to 63 = 216
(2) Starting with 44 and 45 is equal to 36 × 2 = 72
(3) Starting with 433, 434 and 435 is equal to 6 × 3 = 18
(4) Remaining numbers are 4322, 4323, 4324, 4325 is equal to 4
So total numbers are 216 + 72 + 18 + 4 = 310

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Permutations and Combinations
2. Ans. (1)
Since there are 8 males and 5 females. Out of these 13, if we select 11 persons, then there will be at
least 6 males and atleast 3 females in the selection.
13 13 13 × 12
𝑚=𝑛=( )=( )= = 78
11 2 2
3. Ans. (1)
n( n + 1)
+ 99 = ( n − 2)2
2
𝑛2 + 𝑛 + 198 = 2(𝑛2 + 4 – 4n)
𝑛2 – 9𝑛 – 190 = 0
𝑛2 – 19𝑛 + 10 – 190 = 0
𝑛(𝑛 – 19) + 10(𝑛 – 19) = 0
𝑛 = 19
4. Ans. (2)
a1 a2 a3

Number of numbers = 53 – 1
a4 a1 a2 a3

2 ways for a4
Number of numbers = 2 × 53
Required number = 53 + 2 × 53 – 1
= 374
5. Ans. (3)
1
Number of blue lines = Number of sides = 𝑛
𝑛 2
Number of red lines = Number of diagonals = 𝑛 𝐶2 – 𝑛
𝑛
𝑛(𝑛−1)
𝐶2 − 𝑛 = 99 𝑛  − 𝑛 = 99 𝑛 3
2
𝑛−1
− 1 = 99  𝑛 = 201 4
2
6. Ans. (4)
𝐴𝐵𝐶
5 5 5
1 2 2
2 1 2
2 2 1
1 1 3
1 3 1
3 1 1
Total number of selection
= ( 5 𝐶1 5 𝐶2 5 𝐶2 ) × 3 + ( 5 𝐶1 5 𝐶1 5 𝐶3 ) × 3
= 5 × 10 × 10 × 3 + 5 × 5 × 10 × 3 = 2250

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 JEE (Main + Advanced) : Mathematics
7. Ans. (1)
___2_
No. of five digits numbers =
No. of ways of filling remaining 4 places
=8×8×7×6
8 876
 k= =8
336
8. Ans. (309)
MOTHER
1→E
2→H
3→M
4→O
5→R
6→T
So position of word MOTHER in dictionary
2 × 5! + 2 × 4! + 3 × 3! + 2! + 1
= 240 + 48 + 18 + 2 + 1
= 309
9. Ans. (96)

2,4,6,8
4 4 3 2 1
= 4 × 4 × 3 × 2 = 96
10. Ans. (238)
Class 10th 11th 12th
Total student 5 6 8
2 3 5  5 𝐶2 × 6 𝐶3 × 8 𝐶5
Number of selection 2 2 6  5 𝐶2 × 6 𝐶2 × 8 𝐶6
3 2 5  5 𝐶3 × 6 𝐶2 × 8 𝐶5
 Total number of ways = 23800
According to question
100 𝐾 = 23800
 𝐾 = 238
11. Ans. (7744)
209, 220, 231, ..., 495
27
Sum = (209 + 495) = 9504
2
2 3 1
Number containing 1 at unit place 3 4 1
4 5 1
3 1 9
Number containing 1 at 10th place
4 1 8
Required = 9501 – (231 + 341 + 451 + 319 + 418) = 7744

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Permutations and Combinations
12. Ans. (1)
Digits are 1, 2, 2, 3
4!
total distinct numbers = 12 .
2!
total numbers when 1 at unit place is 3.
2 at unit place is 6
3 at unit place is 3.
So, sum = (3 + 12 + 9) (103 + 102 + 10 + 1)
= (1111) × 24
= 26664
13. Ans. (1492)
M A N K I N D

 4  6!   4! 3 
 2!  + (5!  0) +  2!  + (3!  2) + (2!  1) + (1! 1) + (0!  0) + 1 = 1492
   
14. Ans. (1120)
𝑛(𝐵) = 10
𝑛(𝑎) = 5
The number of ways of forming a group of 3 girls of 3 boys.
= 10 𝐶3 × 5 𝐶3
10  9  8 5  4
=  = 1200
3 2 2
The number of ways when two particular boys 𝐵1 of 𝐵2 be the member of group together
= 8 𝐶1 × 5 𝐶3 = 8 × 10 = 80
Number of ways when boys 𝐵1 of 𝐵2 hot in the same group together
= 1200 × 80 = 1120
15. Ans. (7073)
Required no. = Total – no character from {1, 2, 3, 4, 5}
= (106 – 56) + (107 – 57) + (108 – 58) = 106 (1 + 10 + 100) – 56 (1 + 5 + 25)
= 106 × 111 – 56 × 31 = 26 × 56 × 111 – 56 × 31
= 56 (26 × 111 – 31) = 56  7073


  = 7073
16. Ans. (1086)
Let the number is 𝑎𝑏𝑐𝑑, where 𝑎, 𝑏, 𝑐 are divisible by 𝑑.
No. of such numbers
𝑑 = 1, 9 × 10 × 10 = 900
𝑑=2 4 × 5 × 5 = 100
𝑑=3 3 × 4 × 4 = 48
𝑑=4 2 × 3 × 3 = 18
𝑑=5 1×2×2=4
𝑑 = 6, 7, 8, 9 4 × 4 = 16
1086

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 JEE (Main + Advanced) : Mathematics
17. Ans. (1)
7 boys can be seated in 6! ways
now girls will be placed in gaps
 total ways = 6! × 7C5 × 5!
= 126 (5!)2
18. Ans. (72)

Sum of first two digits

Sum of last two digits = 
Case-I :  = 7
2 × 12 = 24 ways.
7 0
0 7 1 6
2 5
3 4
4 3
5 2
6 1
Case-II :  = 8

17 26
71 62
35
53
2 × 8 ways
= 16 ways
Case-III :  = 9

27 36
72 63
45
54
2 × 8 ways
= 16 ways
Case-IV :  = 10

37 46
73 64
2 × 4 ways
8 ways
Case-V :  = 11

47 56
74 65

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Permutations and Combinations
2 × 4 ways
= 8 ways
Ans. 24 + 16 + 16 + 8 + 8 = 72
19. Ans. (4)
Numbers between 5000 & 10000
Using digits 1, 3, 5, 7, 9
5,7,9

3 4 3 2
Total Numbers = 3 × 4 × 3 × 2 = 72
20. Ans. (4)
/M/A/T/H/E/M/A/T/I/
Arrange remaining 9 letters and put 𝐶 and 𝑆 in any 2 gaps out of 10 gaps.
9!
i.e.  10C2  2! = (6!) 𝑘 (Given)
2! 2! 2!
𝑘 = 5670

EXERCISE - JEE (Advanced) PYQ

1. Ans. (5)
When 1 and 2 are removed from numbers 1 to 𝑛 then
we get maximum possible sum of remaining numbers and when 𝑛 –1, 𝑛 are removed then we get
minimum possible sum of remaining numbers.
n( n + 1) n ( n + 1) 
n + n − 2454  0
2
 − ( 2n − 1)  1224  −3   2
2 2 n − 3n − 2446  0

n  50
   n = 50
n  50
Now let 𝑥 and 𝑥 + 1 be two consecutive numbers
50 (50 + 1)
 − x − x − 1 = 1224  𝑥 = 25
2
 25th and 26th cards are removed from pack
 𝑘 = 25  𝑘 – 20 = 5
2. Ans. (7)
as n1  1, n2  2, n3  3, n4  4, n5  5
Let n1 − 1 = x1  0 , n2 − 2 = x2  0 ............ n5 − 5 = x5  0
 New equation will be
x1 + 1 + x2 + 2 + ....... x5 + 5 = 20
 x1 + x2 + x3 + x4 + x5 = 20 − 15 = 5
Now x1  x2  x3  x4  x5

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 JEE (Main + Advanced) : Mathematics
x1 x2 x3 x4 x5
0 0 0 0 5
0 0 0 1 4
0 0 0 2 3
0 0 1 1 3
0 0 1 2 2
0 1 1 1 2
1 1 1 1 1
So 7 possible cases will be there.
3. Ans. (5)
Numbr of red line segments = 𝑛 𝐶2 – 𝑛
Number of blue line segments = 𝑛
 𝑛 𝐶2 – 𝑛 = 𝑛
n ( n − 1)
= 2n  n = 5 Ans.
2
4. Ans. (C)
Total number of dearrangement
1 1 1 1 1
6!  − + − + 
 2! 3! 4! 5! 6! 
= 360 – 120 + 30 – 6 + 1
= 240 + 25 = 265
There are equal chances that card 1 goes into any envelope from 2 to 6
1
 ( 265) = 53
5
5. Ans. (5)
𝑛 = 5!6!
𝑚 = 5! 6C2.5C4.2C1.4!
m
 =5
n
6. Ans. (A)
( 6 𝐶4 + 6 𝐶3 . 4 𝐶1 ). 4 𝐶1 = 380
7. Ans. (5)
𝑥 = 10!
10!
𝑦 = 10 𝐶1 9 𝐶8
2!
𝑦 5.9.10!
= =5
9𝑥 9.10!
8. Ans. (D)
𝑁1 + 𝑁2 + 𝑁3 + 𝑁4 + 𝑁5 = Total ways – {when no odd}
Total ways = 9 𝐶5
Number of ways when no odd, is zero
(∵ only available even are 2, 4, 6, 8)
 9 𝐶5 − zero = 126

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Permutations and Combinations
9. Ans. (625)
Option for last two digits are (12), (24), (32), (44) are (52).
 Total No. of digits = 5 × 5 × 5 × 5 = 625
10. Ans. (119)
𝑛(𝑋) = 5
𝑛(𝑌) = 7
 → Number of one-one function = 7 𝐶5 × 5! 𝑎1 𝑏1
 → Number of onto function 𝑌 to 𝑋 𝑎2 𝑏2
1, 1, 1, 1, 3 1, 1, 1, 2, 2
7! 7!
× 5! + × 5! = ( 7 𝐶3 + 3. 7 𝐶3 )5! = 4 × 7 𝐶3 × 5!
3! 4! (2!)3 3!
𝛽−𝛼 𝑎7 𝑏5
= 4 × 7 𝐶3 −7 𝐶5 = 4 × 35 − 21 = 119
5!
11. Ans. (C)
6 5
(1) 𝛼1 = ( ) ( ) = 200
3 2
So 𝑃 → 4
6 5 6 5 6 5 6 5 6 5
(2) 𝛼2 = ( ) ( ) + ( ) ( ) + ( ) ( ) + ( ) ( ) + ( ) ( )
1 1 2 2 3 3 4 4 5 5
11
=( ) − 1= 46!
5
So 𝑄 → 6
5 6 5 6 5 6 5 6
(3) 𝛼3 = ( ) ( ) + ( ) ( ) + ( ) ( ) + ( ) ( )
2 3 3 2 4 1 5 0
11 5 6 5 6
= ( ) − ( ) ( ) − ( ) ( ) = 381
5 0 5 1 4
So 𝑅 → 5
5 6 4 5 5 6 4 1 5
(4) 𝛼2 = ( ) ( ) − ( ) ( ) + ( ) ( ) − ( ) ( ) + ( )
2 2 1 1 3 1 2 1 4
= 189
So 𝑆 → 2
12. Ans. (30)
When 1𝑅, 2𝐵, 2𝐺 𝐴
5
𝐶1 × 2 = 10 𝐸 𝐵
Other possibilities
1𝐵, 2𝑅, 2𝐺
Or 1𝐺, 2𝑅, 2𝐵 𝐷 𝐶
So total no. of ways = 3 × 10 = 30
13. Ans. (495)
Selection of 4 days out of 15 days such that no two of them are consecutive
15–4+1 12 12×11×10×9
= 𝐶4 = 𝐶4 = 4×3×2
= 11 × 5 × 9 = 495

14. Ans. (1080)

6!
Required ways = × 4! = 1080
2!2!1!1!2!2!

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 JEE (Main + Advanced) : Mathematics
15. Ans. (A,B,D)
(A) 𝑛1 = 10 × 10 × 10 = 1000
(B) As per given condition 1  𝑖 < 𝑗 + 2  10  𝑗  8 & 𝑖  
for 𝑖 = 1, 2, 𝑗 = 1, 2, 3, ..., 8 → (8 + 8) possibilities
for 𝑖 = 3, 𝑗 = 2, 3, ..., 8 → 7 possibilities
𝑖 = 4, 𝑗 = 3, ..., 8 → 6 possibilities
𝑖 = 9, 𝑗=1 → 1 possibility
So 𝑛2 = (1 + 2 + 3 + .....+ 8) + 8 = 44
(C) 𝑛3 = 10 𝐶4 (Choose any four)
= 210
(D) 𝑛4 = 10 𝐶4 . 4! = (210) (24)
n
 4 = 420
12
So correct Ans. (A), (B), (D)
16. Ans. (569.00)
2,3,
(1) 2 0 2 4,6,7 5

(2) 3,4,
2 0 6,7 24

4 6
2,3,4
(3) 2 ,6,7
180

5 6 6

(4) 3 216

6 6 6
0,2
(5) 4 ,3,4 144

4 6 6
Number of 4 digit integers in [2022,4482]
= 5 + 24 + 180 + 216 + 144 = 569
17. Ans. (A)
3R 3R 3R 3R
2B 2B 2B 2B
𝐵-1 𝐵-2 𝐵-3 𝐵-4
Case-I : when exactly one box provides four balls (3𝑅 1𝐵 or 2𝑅 2𝐵)
Number of ways in this case 5 𝐶4 ( 3 𝐶1 × 2 𝐶1)3 × 4
Case-II : when exactly two boxes provide three balls (2𝑅 1𝐵 or 1𝑅 2𝐵) each
Number of ways in this case ( 5 𝐶3 – 1)2 ( 3 𝐶1 × 2 𝐶1)2 × 6
Required number of ways = 21816
Language ambiguity : If we consider at least one red ball and exactly one blue ball, then required
number of ways is 9504. None of the option is correct.

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Permutations and Combinations
18. Ans. (31)
No. of elements in 𝑋 which are multiple of 5
4 
____ 0 → =4  4 
3 ____ 0 → =4 
1,2,2,2 fixed  fixed
3


1,2,2,2
4 
____ 0 → = 12  ____ 0 →
4
= 12 
1,4,2,2 fixed
2  2 
 1,4,2,2 fixed

= 4  Total = 38
4
= 4  Total = 38
4
____ 0 → ____ 0 →
3 3
4,2,2,2 fixed  4,2,2,2 fixed 
 4 
4 ____ 0 → = 6
____ 0 → = 6
22  2,2,4,4 fixed
22 
2,2,4,4 fixed 

= 12 
4
____ 0 →
= 12 
4
____ 0 → 2 
2  1,2,4,4 fixed 
1,2,4,4 fixed 
Among these 38 elements, let us calculate when element is not divisible by 20
3 
___ 1 0 → =1
3 
2,2,2 fixed 
3 
_ _ _ 1 0 → = 3 Total = 7
2
2,2,4 fixed 
3 
_ _ _ 1 0 → = 3
2 
2,4,4 fixed 
38 − 7
 p=  38𝑝 = 31
38

JEE (Main) Practice Paper

SECTION–A
1. Ans. (3)
𝐷 𝐶

𝐸
𝐴 𝐵
𝐴, 𝐸, 𝐶 are collinear
& 𝐵, 𝐸, 𝐷 are collinear
 5𝐶 3 – 2.3𝐶 3 = 8
2. Ans. (2)
Teacher's visits = 25𝐶 5 (different sets of all kids) A kid's visit = 24𝐶 4 (different sets of other kids)
Difference = 25𝐶 5 – 24𝐶 4.
= 24𝐶 5 { n +1 Cr – n Cr −1 = n Cr }
3. Ans. (3)
n
C3 – [𝑛 + 𝑛(𝑛 – 4)] = 30
𝑛 = 9 satisfies it

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4. Ans. (3)
A ................. = 4! = 24
N ................= 4! = 24
R ............... = 4! = 24
U ............... = 4! = 24
VAN .......... = 2! = 2
VARN ........ = 1
VARUN .......= 1
VARUN = 100
5. Ans. (2)
If last digit is 5 5 = 2688
8× 8× 7×6

If Last digit is 0 choice of places for 5 = 4  0 = 1344

4× 8× 7×6
Total = 4032  𝑘 = 168
6. Ans. (4)
Select any 4 pairs and one shoe from each pair
= 5C4 . 2C1 . 2C12C12C1 = 80
7. Ans. (2)
Consider 𝐴 & 𝐵 as one. 𝐶 & 𝐷 are not to be arranged.
7! 2! → To arrange A & B
Total =
2! → For not arranging C & D
= 5040
8. Ans. (3)
1, 2, 3, 4, 5, 0
A number is divisible by 25 if the last two digits are 25 or 50
(i) Now, if 5 is not taken then number of such numbers = 0
(ii) If 0 is not taken
2 5 =6
(iii) if 2 is not taken then
5 0 = 6 and
(iv) If 1 | 3 | 4 is rejected
then in each case we have,
5 0 =6

2 5 =4
2.2 1
if 3, 4, or 1 is not taken then 10 number for each case.
Total = 30
Hence, Total = 30 + 6 + 6 = 42

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Permutations and Combinations
9. Ans. (1)
Last match is won by India.
0 match won by Pakistan = 4𝐶 0 = 1
1 match won by Pakistan = 5𝐶 1 = 5
2 matches won by Pakistan = 6𝐶 2 = 15
3 matches won by Pakistan = 7𝐶 3 = 35
4 matches won by Pakistan = 8𝐶 4 = 70
Total = 126
10. Ans. (1)
1, 2, will be arranged in only one ways and similarly 3, 4, 5, 6 will be arranged in only one way.
9!
Number of permutations = 9.7!
2!2!2!
11. Ans. (3)
5 in car I & 3 in car II = 8 C5
4 in car I & 4 in car II = 8 C4
Total = 56 + 70 = 126
12. Ans. (3)
p+q
L= 1
p q
𝑥 gets 𝑝 books and 𝑦 gets 𝑞 books
p+q
M= 2
p q
p+q
N=
p q
clearly 2𝐿 = 𝑀 = 2𝑁
13. Ans. (4)
(1 3 4) or (2 2 4) or (2 3 3)
 8! 8! 8! 
= + +   3! = (280 + 210 + 280) × 3!
 1! 3! 4! 2! 2! 4! 2! 2! 3! 3! 2! 
770  6
𝑘= = 22
765
14. Ans. (2)
Any one subjective is allotted two periods
6!
= 5𝐶 1 × = 1800
2!
15. Ans. (3)
(𝑥 1 + 𝑥 2 + 𝑥 3) (𝑦1 + 𝑦2) = 11 · 7 or 7 · 11
In the first case (𝑥 1 + 𝑥 2 + 𝑥 3) = 11 and (𝑦1 + 𝑦2) =7, which have 10𝐶 2·6𝐶 1 solutions (using beggar)
In the second case (𝑥 1+𝑥 2 + 𝑥 3) = 7 & (𝑦1 + 𝑦2) = 11, which have 6𝐶 2 · 10𝐶 1 solutions (using beggar)
 total number of solutions = 10𝐶 2 · 6𝐶 1 + 6𝐶 2 · 10𝐶 1 = 270 + 150 = 420 Ans.

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 JEE (Main + Advanced) : Mathematics
16. Ans. (3)
Let 𝑖 appears on 𝑎𝑖 dice 𝑖 = 1, 2, 3, 4, 5, 6
so no. of out comes is equal to no. of solution of a1 + a2 + a3 + a4 + a5 + a6 = n = ( n + 5) c5
17. Ans. (1)
Required number of ways 36𝐶 9 . 27𝐶 9 . 18𝐶 9 . 9𝐶 9 . 4 ! = 36!4 × 4!
(9!)
18. Ans. (2)
If 𝑃3 sits on 𝐶 1
4!  1 – 1 + 1 – 1 + 1  = 4.3 – 4 +1 = 9
 1! 2! 3! 4! 
If P3 does not sit on C1
= 5!  1 – 1 + 1 – 1 + 1 – 1  = 44
 1! 2! 3! 4! 5! 
total number of ways = 44 + 9 = 53
 P1 C1
P C →
 2 2 P1
 P3 C3 → P2

 P4 C4
P C
 5 5

 P6 C6

 P7 C7
19. Ans. (1)
First we select 3 length from the given 6 length so the no. of ways = 6𝐶 3
But these some pair i.e. (2, 3, 7), (2, 3, 6), (2, 3, 5) (2, 4, 6), (2, 4, 7), (2, 5, 7), (3, 4, 7) are not form a
triangle so that total no. of ways is
6𝐶 3 – 7 ways

20. Ans. (1)

𝑚, 𝑛 both product of element of one of the subset of {222, 310, 56, 73, 112, 13, 17, 19, 23} such that
𝑚𝑛 = 25!  number of ways of selecting '𝑚' is 29 ways (here 𝑛 is automatically fixed according to m)
29
 Total number of required ways = = 256
2
m m
{because in half of the ways > 1 and in half of the ways 0 < < 1}
n n

SECTION–B
1. Ans. (93)
Number divisible by 3 if sum of digits divisible
case-I If 1 + 2 + 3 + 4 + 8 = 18 Number of ways = 120
case-II If 1 + 2 + 3 + 7 + 8 = 21 Number of ways = 120
case-III If 2 + 3 + 4 + 7 + 8 = 24 Number of ways = 120
case-IV If 1 + 2 + 0 + 4 + 8 = 15 Number of ways = 96
case-V If 1 + 2 + 0 + 7 + 8 = 18 Number of ways = 96
case-VI If 2 + 0 + 4 + 7 + 8 = 21 Number of ways = 96
case-VII If 0 + 1 + 3 + 4 + 7 = 15 Number of ways = 96
total number 744

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Permutations and Combinations
2. Ans. (41)
Total number of triangle = Two points taken from 𝐴𝐵 and one point either 𝐵𝐶 or 𝐶𝐴 + similarly
𝐵𝐶 + similarly 𝐴 + one point each sides. 𝐶
= 3𝐶 2 [4𝐶 1 + 5𝐶 1] + 4𝐶 2 [5𝐶 1 + 3𝐶 1] + 5𝐶 2 [3𝐶 1 + 4𝐶 1] + 3𝐶 14𝐶 15𝐶 1 = 205
Total – (collinear points used)
= 12𝐶 3 – (3𝐶 3 + 4𝐶 3 + 5𝐶 3) = 220 – 15 = 205
Alternate 𝐴 𝐵
Total – Collinear points used = 12𝐶 3 – (3𝐶 3 + 4𝐶 3 + 5𝐶 3) = 220 – 15 = 205
3. Ans. (63)
Coin dividing in any are possible i.e.
1, 2, 4
so the number of ways is
7𝐶 1 . 6𝐶 2 . 4𝐶 4 . 3! = 7 × 15 × 6 = 630

4. Ans. (62)
Let number be x1 x2 x3 x4 x5 x6
But Here x1 + x2 + .... x6 = 12

( ) = (1 − x ) . (1 − x ) –6
6 6
So coefficient of 𝑥 12 in expansion 1 + x + x2 + .... + x9 10

 17C12 − 6C1 . 7C2 = 6188 – 126 = 6062

5. Ans. (10)
Let team 𝑋 wins '𝑚' matches, if it wins (𝑚 + 𝑟)𝑡ℎ match and wins 𝑚 – 1 match from the first
20
m Cm
𝑚 + 𝑟 – 1 matchs, so total no. of ways =  m + r −1
C m−1 = hence 𝑚 = 10
r =0 2
6. Ans. (42)
SERIES
S - 2, E - 2 , R, 
case-I when all letter distinct is
4𝐶 3 × 3! = 4 × 6 = 24

case-II when 2 letters are same the

2𝐶 1 . 3𝐶 1 ×
3!
= 2 . 3 . 3 = 18
2!
total number is 24 + 18 = 42
7. Ans. (31)
Case -I If all are different then no. of ways is = 6𝐶 3 = 20
Case-II If three each of two colours, then combination is
3 0 → 2!
2 1 → 2! = 2! + 2! = 4 ways
Case-III If two each of three colours, then combination is
2 1 0 → 3!
1 1 1 → 1! = 3! + 1! = 7ways
Hence required no.is = 20 + 7 + 4 = 31
8. Ans. (13)
Using multinomial theorem total number of required selection is 8+3𝐶 8 = 11𝐶 8 = 11𝐶 3 = 165
9. Ans. (28)
𝑃𝑛 = 𝑛–2 𝐶3
Pn+1 − Pn = n–1C3 − n–2C3 = n–2C2 = 15
 (𝑛 – 2)(𝑛 – 3) = 30  𝑛 = 8

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 JEE (Main + Advanced) : Mathematics
10. Ans. (1)
2002 (2002)!
C1001 =
(1001)!(1001)!
no. of zeros in (2002)! are
400 + 80 + 16 + 3 = 499
no. of zeroes in (1001 !)2 = 2(200 + 40 + 8 + 1) = 498
Hence no. of zeroes is (2002)!2 = 1
(1001!)

JEE (Advanced) Practice Paper

1. Ans. (C)
11
C3 + 11C4 + ....... + 11C11 = 211 − 11C0 − 11C1 −11 C2 = 1981
2. Ans. (B)
–
= y1 y2 y3
– x1 x2 x3
__________
Number of pairs = (10 + 9 + 8 +….. + 1)2 . (9 + 8 +……. + 1) = (55)2 .45
3. Ans. (D)
Total 𝑛-digit numbers using 1, 2 or 3 = 3n
total 𝑛-digit numbers using any two digits out of 1, 2 or 3 = 3𝐶 2 × 2n – 6 = 3 × 2n – 6
total 𝑛-digit numbers using only one digit of 1, 2 or 3 = 3
 the numbers containing all three of the digits
1, 2 and 3 at least once = 3n – (3 × 2n – 6) – 3 = 3n – 3 . 2n + 3
4. Ans. (A)
Ordered pair = total – (𝐴  𝐵 = 𝑋) = 4n − 3n
Subsets of X = 2n will not repeat in both but here the whole set 𝑋 has not been taken
So subsets of 𝑥 which are not repeated 2n − 1 ( )
(4n − 3n ) − (2n − 1)
Hence unordered pair = + (2n – 1)
2
5. Ans. (C)
Using multinomial theorem
17  16 12  11
Total no. of ways = 15 + 3 – 1𝐶 15 × 10 + 3 – 1𝐶 10 = 17𝐶 15 × 12𝐶 10 =  = 8976
2 2
6. Ans. (D)
𝐴 𝐵 𝐴 𝐵 𝐴 𝐵 = total number of required possible is

𝐵 𝐴 𝐵 𝐴 𝐵 A

12𝐶 6
12!
× 6 ! × 6! × 2! = × 6 ! × 6 ! × 2! = 2 × 12!
6! 6!

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Permutations and Combinations
7. Ans. (B,C)
Total required number of teams is
1
= 10𝐶 4 · 6𝐶 3 · 3𝐶 3 · 2100 = 10𝐶 4 · 5𝐶 2 = 2100
2!
8. Ans. (C,D)
SAMANVAYA
 8!
(9 − 1)! 8!  0!4!
Number of circular permutations is = = =
4! 4!  8!
 1!4!
0 + 16 − 8 = 8
x2 + y 2 − n = 
1 + 16 − 8 = 9
9. Ans. (A,B,D)
.........08 
.........60 →  3  2  1  3 = 18 + = 30
.........80 → 
.........16 → 
.........36 →  2  2  1  3 = 12
.........68 → 
Similarly B & D are also correct
10. Ans. (B,D)
Number of 7 digit numbers with starting digit 1 will be 6! = 720. But these also contain numbers
ending at the digit 5, which become divisible by 5. So, fixing starting digit 1 and ending digit 5, we
get 5! = 120 numbers. Thus, number of 7 digit numbers starting with 1 and not divisible by 5 will
be 720 – 120 = 600.
Similarly, number of 7 digit numbers starting with 2 and not divisible by 5 will be 600. Similarly,
number of 7 digit numbers starting with 3 and not divisible by 5 will also be 600.
Thus, when put in required increasing order, 1801𝑡ℎ number will be the smallest numbers starting
with 4 and not divisible by 5 which is 4123567.
Similarly we can check other options also
11. Ans. (B,C)
a2 − b2 = c ( a − b )( a + b ) = c
 𝑎 – 𝑏 = 1  𝑎 = 3, 𝑏 = 2, 𝑐 = 5 & 𝑑 = 7
and sum of all possible numbers
= 3!(1111)(2 + 3 + 5 + 7) = 113322
12. Ans. (A,B,C)
 31   31   31   31 
Exponent of 2 in 31! =   +  2  +  3  +  4 
 2  2  2  2 
= 15 + 7 + 3 + 1 = 26
Exponent of 2 in 33! = 31
Exponent of 2 in 35! = 32
So 31! = 226 1
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 JEE (Main + Advanced) : Mathematics
33! = 2312
35! = 2323
 maximum value of 𝑚 is 89
13. Ans. (C)
𝑃1 , 𝑃2 , 𝑃3 , 𝑃4
16
12 others
(12)! (12)·(11)! (11)!
12 others can be divided into 4 equal groups in each of 3 person = =
(3!)4 6·6·6·6 108
14. Ans. (D)
𝑃1 , 𝑃2 , 𝑃3 , 𝑃4 3
10!
16 𝑃6 ; 10 =
3! 7!
𝑃7 − 𝑃16 (10) 7
000 𝑥𝑥𝑥𝑥𝑥𝑥𝑥
+𝑃6 𝑃1 to 𝑃5
4 players 12 players
4
12!
Now, 12 4 =
(4!)3 ·3!
4
(10)! (12)!  12! 
Total ways = . =  .20
3! 7! (4!)3 ·3!  (4!)3 
 𝑘 = 20
15. Ans. (C)
𝑁 = 2910600
N = 23.33.52.72.11
Divisible by (15)
𝑁 = (5.(3))(23.32.5.72.11)

(15)
(23 ) (32 ) (5) (72 ) (11)
N
=     
15
(3 + 1) (2 + 1) (1 + 1) (2 + 1) (1 + 1)
 144
Divisible by 180 → (L.C.M. of 15 and 36)
𝑁 = (22.32.5) (2) (3) (5) (72) (11)
(2) (3) (5) (7) (11)
N
=     
180
(1 + 1) (1 + 1) (1 + 1) (2 + 1) (1 + 1)
 48
Divisible by 15 not by 36
 144 – 48 = 96

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Permutations and Combinations
16. Ans. (C)
𝑁 = 2333527211
for H.C.F. = 2
7* → 2 choice
11* → 2 choice
 2 × 2 × 2 × 2 = 16 ways
same for H.C.F. = 3 = 16 ways
but for H.C.F. = 5 or 7
same for H.C.F. = 3 = 16 ways
But for H.C.F. = 5 or 7
for first pocket (3*, 2*, 7* or 11*) we only have one choice and for rest we have 2 choice
 2 × 2 × 2 × (1) = 8
Total  16 + 16 + 8 + 8  48
(HCF 11 not possible)
17. Ans. (B)
( A1 , A2 ) , ( B1 , B2 ) , (C1 , C2 ) , D, E , F , G
we select 2 pairs and keep them together and arrange the remaining pair within gaps after
arranging together pair and singles
= 3 C2  (6 − 1)!  6 C2  2 2 2 = 60  6!
2 pairs Circular 2 gaps Arranging
chosen arrangement out of pairs at
of 2 pairs and si x for respective
and sin gles places .

18. Ans. (A)

A1 A2 B1 B2C1C2 DEFG
Only A1 , A2 = Together & way sin − which
other pairs separat A1 , A2 together
 
a s
     
 A1 A2together B1 B2together   A1 A2together , C1C2together   A1 A2together , B1 B2together 
     
 but C1C2 separate   & B1 B2 separate + C1C2 together 
−
        
     
 b   c   d 
     
𝐴𝐴 𝐵𝐵

𝐶𝐶

𝑠 = 8! 2!
𝑏 = 7! × 2! 2! = c
𝑑 = 6! 2! 2! × 2!
 𝑎 = 8!2! – (7! × 2! × 2! × 2) + 6!2!2!2!
= 6! (2!(8 × 7 × 7 × 2 × 2 + 4)
 𝑎 = (64) 6!  𝑞 = 64

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