Level-I

Chapter 10

Permutations and Combinations

Solutions (Set-1)

Very Short Answer Type Questions :

1. A coin is tossed n times. Find the total number of possible outcomes.
Sol. When a coin is tossed, the number of possible outcomes is two i.e., either head or tail.

∴ Required number of possible outcomes = 2n

1 1 x
2. If + = , then find x.
8! 9! 10!

1 1 x
Sol. + =
8! 9! 10!

9 1 x
⇒ + =
8! × 9 9! 10 × 9!

9 1 x
⇒ + =
9! 9! 10 × 9!

10 x
⇒ =
9! 10 × 9!

∴ x = 100

3. Find n, if (n + 1)! = 12. n!.

Sol. (n + 1)! = 12 n!

(n + 1)!
⇒ = 12
n!

(n + 1) n !
⇒ = 12
n!

∴ n = 11

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138 Permutations and Combinations Solutions of Assignment (Level-I) (Set-1)

4. Prove that, 2.C(7, 4) = C(8, 4) where C.(n, r) = nCr

Sol. LHS = 2.C(7, 4)
7!
= 2×
4! 3!
2 × 4 × 7!
= [Multiplying 4 in Numerator and denominator]
4 × 4! × 3!
8 × 7! 8!
= = = C(8, 4) = RHS
4 × 3! × 4! 4! 4!
Hence proved.
5. If 8Cr – 7C3 = 7C2, find the value of r.
Sol. 8Cr – 7C3 = 7C2
⇒ 8C
r = 7C3 + 7C2
⇒ 8C
r = 8C3 [∵ nCr + nCr – 1 = n + 1C ]
r
∴ r=3
6. In how many ways the letters of the word EAMCOT can be arranged?
Sol. Number of letters in the word EAMCOT = 6
∴ Number of arrangements of the letters of the word EAMCOT = 6P6 = 6! = 720
7. In how many ways can 7 students be selected from 12 students?
Sol. Number of selections of 7 students out of 12 = 12C = 792
7
8. How many triangles can be formed by joining nine points when no three of them are collinear.
Sol. Triangles are formed when 3 non-collinear points are joined.
Hence, number of triangles = 9C3 = 84
9. Find the unit digit in the expansion of 1! + 2! + 3! + 4!.
Sol. 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33
∴ Unit digit is 3

Short Answer Type Questions :

10. In how many ways can a student choose 7 subjects out of 10 subjects if 3 subjects are compulsory?
Sol. 3 subjects are compulsory, it means 4 subjects have to chosen out of the remaining 7 subjects. This can be
done in 7C4 ways.
Hence, number of selections = 7C4 = 35
11. In how many ways can a football team of 11 players be selected from 16 players? In how many of these will include
2 particular players?
Sol. Number of selections of 11 players out of 16 can be done in 16C ways.
11
If 2 particular players is always included, it means remaining 9 players can be selected out of the remaining
14 players.
Hence, number of required selection = 14C
9
12. There are n points on a circle. Find the number of straight lines formed by joining them.
Sol. A straight line is formed when any two points are joined to each other.
Number of points on the circle = n
n(n − 1)
∴ Number of straight lines = nC2 =
2

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Solutions of Assignment (Level-I) (Set-1) Permutations and Combinations 139

( x + 2)! (2 x + 1)! 72
13. If x ∈ N, solve the equation : . = .
(2 x − 1)! ( x + 3)! 7
( x + 2)! (2 x + 1)! 72
Sol. × =
(2 x − 1)! ( x + 3)! 7
(2 x + 1)(2 x ) 72
⇒ =
( x + 3) 7
⇒ 7x(2x + 1) = 36 (x + 3)
⇒ 14x2 + 7x – 36x – 108 = 0
⇒ 14x2 – 29x – 108 = 0

−b ± D
⇒ x=
2a
29 ± 83
⇒ x=
28
54
⇒ x = 4,
28
−54
But, x ≠
28
∴ x=4
14. In how many ways can the letters of the word DELHI be arranged so that the letter E and H occupy only even
places?
Sol. In the word DELHI, there are two even place which can arranged by the letter E and H in 2! ways. Remaining
three letters can be arranged in remaining three odd places in 3! ways.
∴ Required number of arrangements = 3! × 2!
= 12
15. Six men and five women are to sit in a row so that the women occupy the even places. Find the number of all
possible arrangements?
Sol. Total number of persons is 11 (six men and five women). Out of these there are 5 even places in which 5
women can be arranged in 5! ways. In six odd places out of 11 places 6 men can arranged in 6! ways.
∴ Required number of arrangements = 5! × 6!
= 120 × 720 = 86400
16. Rajeev wants to arrange 3 Economics, 2 History and 7 English books on a shelf. If the books on the same subject
are different. Find the number of possible arrangements if all the books on a subject are together.
Sol. Since books of the same subjects are to be put together.
∴ Number of arrangements of 3 Economics books = 3! = 6
Number of arrangements of 2 History books = 2! = 2
Number of arrangements of 7 English books = 7! = 5040
Set of books can be arranged in 3! ways
∴ Total number of arrangements of all books, when books on a same subject are to be together
= 6 × 2 × 5040 × 6
= 60480 × 6
= 362880

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140 Permutations and Combinations Solutions of Assignment (Level-I) (Set-1)

17. If nP5 = 60.n – 1P3, then find the value of n.

Sol. nP5 = 60 n – 1P
3
n
P5
⇒ n −1
= 60
P3

n ! (n − 4)!
⇒ = 60
(n − 5)! (n − 1)!

⇒ n(n – 4) = 60
⇒ n(n – 4) = 6 × 10
∴ n = 10
18. Find the number of ways in which the letters of the word ARRANGE can be permuted such that R’s occur together.
Sol. In the word ARRANGE, there are two R. Let two R be as a single letter therefore, remaining letters in the
word ARRANGE is 6, out of these
A occurs twice and all are different

6!
∴ Number of arrangements = = 360
2!

2!
Now, 2R can be arranged in = 1 way
2!

∴ Total number of arrangements of the word ARRANGE, in which R’s occurs together is 360
19. Find the number of all five digit numbers that can be formed by using the digits 0, 1, 3, 5, 6.
Sol. For a five digit numbers, 0 cannot be put at ten thousands place, Hence, available choices to fill ten thousands
place is 4. Now the remaining places can be filled by any of the 5 digits.
∴ Required number of arrangements = 4 × 54
= 4 × 625 = 2500
20. How many words can be formed out of the letters of the word PECULIAR beginning with P and ending with R?
Sol. Number of letters in the word PECULIAR = 8. P and R are fixed at left end and right end respectively.
∴ Number of remaining letters = 6
∴ Required number of words = 6! = 720
21. In how many ways can the letters of the word FRACTION be arranged so that no two vowels are together?
Sol. In the word FRACTION, there are 3 vowels (A, I, O). Let it be a set of single letter, remaining letter in the
word FRACTION is 6, which can be arranged themselves in 6! ways. Again the set of vowels (A, I, O) can
be arranged in 3! ways.
∴ Number of arrangements of letters of the word FRACTION in which vowels are together = 6! × 3!
Number of arrangements of letter of the word FRACTION = 8!
∴ Required number of arrangements = Number of arrangements of letter of the word FRACTION – Number
of arrangements of letters of the word FRACTION in which vowels are together
= 8! – 6! × 3!
= 6! (7 × 8 – 6)
= 720 × 50 = 36000

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Solutions of Assignment (Level-I) (Set-1) Permutations and Combinations 141
22. Determine the number of 6 card combinations out of a deck of 52 cards if two of the six cards have to be aces.
Sol. Total number of cards = 52
Number of aces = 4
Remaing number of cards = 48
Required number of combinations of 6 cards = 4C2 × 48C
4
23. Out of six consonants and four vowels, how many words each containing 3 consonants and 2 vowels can be formed?
Sol. 3 consonants can be selected out of 6 in 6C3 ways
2 vowels can be selected out of 4 in 4C2 ways
∴ Number of required selection = 6C3 × 4C2
Now, the group of 5 letters can be arranged in 5! ways
∴ Required number of words = 6C3 × 4C2 × 5!
= 20 × 6 × 120
= 14400
24. The letters of the word RACHI are arranged in all possible ways as listed in dictionary. What is the rank of the
word RACHI?
Sol. In dictionary the words at each stage are arranged in alphabetical order. In this problem we must consider
the words beginning with A, C, H, I, R in order.
∴ Number of words starting with A = 4! = 24
Number of words starting with C = 4! = 24
Number of words starting with H = 4! = 24
Number of words starting with I = 4! = 24
Number of words beginning with R is 4!, but one of these words is the word RACHI. But RACHI is the first
word starting with R.
∴ Rank of RACHI = 4 × 24 + 1 = 97
25. Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the number
of lines that can be formed by joining the points.
Sol. Total number of points = 18
Number of collinear points = 5
A line is formed when two points are joined together
∴ Required number of lines = 18C
2 – 5C2 + 1
= 153 – 10 + 1
= 144
26. How many committees of five persons and a chairperson can be selected from 15 persons?
Sol. Chairperson can be selected in 15 ways.
Now the remaining 5 persons of the committee out of 14 persons can be chosen in 14C ways.
5
∴ Required number of selections = 15 × 14C
5

Long Answer Type Questions :

27. A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be
drawn from the bag, if
(i) They can be of any colour.
(ii) All must be of the same colour.

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142 Permutations and Combinations Solutions of Assignment (Level-I) (Set-1)

Sol. (i) If balls are of any colour, it means 4 balls are drawn from total of 11 balls and this can be done in 11C
4
ways.
11C = 330
4

(ii) If all are of same colours, then it may be carried in following ways:
(a) Drawing of 4 red balls
(b) Drawing of 4 white balls
Drawing of 4 red balls out of 5 is done in 5C4 ways.
Drawing of 4 white balls out of 6 is done in 6C4 ways.
The job is performed in either of the way.
∴ Required number ways = 6C4 + 5C4
= 15 + 5
= 20
28. A group consists of 4 girls and 6 boys. In how many ways can a team of 4 members be selected, if the team
has
(i) No girls
(ii) At least two boys and one girl.
Sol. (i) If no girls is selected in a team of 4 members. It means 4 boys are selected out of 6 boys.
∴ Number of required selections = 6C4
= 15
(ii) 4 members of team of at least two boys and one girl can be selected in the following way:
(a) One girl and three boys
(b) Two girls and two boys
Number of selections of one girl and three boys = 4C1 × 6C3
= 4 × 20 = 80
Number of selections of two girls and two boys = 4C2 × 6C2
= 6 × 15 = 90
Team can selected in either of the above two ways
∴ Total number of selections = 80 + 90
= 170
29. (i) How many triangles can be formed by joining the vertices of a nanogon?
(ii) How many diagonals are there in a polygon with 8 sides?
Sol. (i) A nanogon has 9 sides. A triangle is formed when three points are joined in pair wise
∴ Number of triangles = 9C3
= 84
(ii) Number of diagonals of a polygon having n sides is given by nC2 – n
Number of sides of the polygon = 8
∴ Number of diagonals = 8C2 – 8
= 28 – 8 = 20
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Solutions of Assignment (Level-I) (Set-1) Permutations and Combinations 143
30. If n + 1C : nCr : n – 1C = 11 : 6 : 3, then find the values of n and r.
r+1 r–1

Sol. n + 1C : nCr : n – 1C = 11 : 6 : 3
r+1 r–1

n +1
Cr + 1 11
Now, n
=
Cr 6

n + 1 11
⇒ =
r +1 6
⇒ 6n – 11r = 5 …(i)

n
Cr 6
Again, n −1
=
Cr − 1 3

n
⇒ =2
r
⇒ n = 2r …(ii)
On solving equations (i) and (ii), we get
n = 10 and r = 5
31. If m + nP = 90 and m – nP = 30, then find (m, n).
2 2

Sol. m + nP = 90 and m – nP = 30
2 2

(m + n )! (m − n )!
⇒ = 90 and = 30
(m + n − 2)! (m − n − 2)!

⇒ (m + n) (m + n – 1) = 90 and (m – n) (m – n – 1) = 30
Now, (m + n) (m + n – 1) = 10 × 9 and (m – n) (m – n – 1) = 6 × 5
⇒ m + n = 10 …(i) and
m–n=6 …(ii)
An solving equations (i) and (ii), we get
m = 8 and n = 2
∴ (m , n) = (8, 2)
32. If nPr = nPr + 1 and nCr = nCr – 1, find n and r.
Sol. nPr = nPr + 1 and nCr = nCr – 1
n
n
Pr Cr
⇒ = 1 and n
=1
n
Pr + 1 Cr − 1
 n
Cr n − r + 1
n ! (n − r − 1)! n − r +1 ∵ = 
⇒ = 1 and =1 n
Cr − 1 r
(n − r )! n ! r  
1
⇒ = 1 and n – r + 1 = r
n−r
⇒ n–r=1 …(i) and
n – 2r = –1 …(ii)
An solving equations (i) and (ii), we get
n = 3 and r = 2

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144 Permutations and Combinations Solutions of Assignment (Level-I) (Set-1)

33. If n + 2C : n – 2P = 57 : 16, prove that n = 19.

8 4

n+2
C8 57
Sol.
n−2
=
P4 16

(n + 2)! (n − 6)! 57
⇒ =
8! (n − 6)! (n − 2)! 16

(n + 2)! 57
⇒ =
8 × 7! (n − 2)! 16

(n + 2)(n + 1)n(n − 1)(n − 2)! 57 × 7!

⇒ =
(n − 2)! 2

⇒ (n + 2) (n + 1)n (n – 1) = 57 × 7 × 6 × 5 × 4 × 3
⇒ (n + 2) (n + 1)n (n – 1) = 21 × 20 × 19 × 18
∴ n = 19
Hence proved.
34. In how many ways can the letters of the word INTERMEDIATE be arranged so that
(i) The vowels always occupy even places?
(ii) The relative order of vowels and consonants do not alter?
Sol. (i) Number of letters in the word INTERMEDIATE = 12. Out of these 6 places are even and six places are
odd. Number of vowels are six out of these E occurs thrice, I occurs twice and A occurs only one.

6!
Now, number of arrangements of vowels = = 60
3! 2!

In remaining six odd position six consonants can be arranged in which T occurs twice and all are different.

6!
∴ Number of arrangements of consonants = = 360
2!
Hence, required number of arrangements = 360 × 60 = 21600
(ii) Number of vowels = 6 and
Number of consonants = 6
Since, relative order of vowels and consonants cannot be changed. It means 6 vowels can be arranged
in their 6 relative positions as well as 6 consonants can be arranged in their 6 relative positions.

6! 6!
∴ Required number of arrangements = × = 360 × 60 = 21600
2! 3! 2!

35. If the letters of the word RELATE be permuted and the words so formed be arranged as in a dictionary, find the
rank of the word RELATE.
Sol. In dictionary the words at each stage are arranged in alphabetical order starting with the letter A, E, L, R and T.

5!
∴ Number of words starting with A = = 60
2!
Number of words starting with E = 5! = 120
5!
Number of words starting with L = = 60
2!

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Solutions of Assignment (Level-I) (Set-1) Permutations and Combinations 145

5!
Number of words starting with R is = , but one of these words is the word RELATE.
2!

4!
∴ Number of words starting with RA = = 12
2!
Number of words starting with REA = 3! = 6
Number of words starting with REE = 3! = 6
Now, the first word beginning with REL is the word RELAET and the next word is RELATE.
∴ Rank of RELATE = 120 + 2 × 60 + 12 + 2 × 6 + 2 = 266
36. How many words can be formed by taking 4 letters at a time out of the letters of the word EXAMINATION?
Sol. Number of letters in the word EXAMINATION is 11, out of these A occurs twice, I occurs twice, N occurs twice
and all are different. Now 4 letters can be choosed in following ways.
(i) All the four distinct letters : There are 8 distinct letters out of these, 4 can be chosen in 8C4 ways.
Now, 4 different letters can be arranged themselves in each selection in 4! ways.
Hence, number of words = 8C4 × 4! = 1680
(ii) Two distinct and two alike letters : There are 3 pairs of alike letters out of which one pair can be
chosen in 3C1 ways. Now, the two distinct letters out of the remaining 7 letters can be chosen in 7C2
ways. Therefore, number of selections of 4 letters is 7C2 × 3C1. The 4 letters can be arranged in each
4!
selection, in ways.
2!

4!
Hence, number of words = 7C2 × 3C1 × = 756
2!
(iii) Two alike of one kind and two alike of other kind : There are 3 pairs of 2 alike letters out of these
4!
2 pairs can be chosen in 3C2 ways. In each of these selection 4 letters can be arranged in ways.
2! 2!

4!
Hence, number of words = 3C2 × = 18.
2! 2!

∴ Total number of 4 letter words = 1680 + 756 + 18 = 2454

37. How many words with three different vowels and three different consonants can be formed from the English alphabet?
Sol. Number of alphabets is 26, out of these 5 are vowels and the remaining 21 are consonants. Now 3 vowels
can be selected in 5C3 ways and the three consonants can be selected in 21C3.
∴ Number of selections of 3 vowels are 3 consonants = 5C3 × 21C .
3 In each of these selection of 6 letters,
letters can be arranged in 6! ways.
∴ Total number of words = 5C3 × 21C
3 × 6!
38. In how many ways can the letters of the word PENCIL be arranged so that
(i) N is always next to C?
(ii) N and E are always together?
Sol. (i) Let N and C as a single letter in which N is next to C. Remaining letters in the word PENCIL after taking
N and C as a single letter is 5, which can be arranged in 5! ways.
Hence, required number of words = 120
(ii) By taking N and E as a single letter. Now, word PENCIL is left with 5 letters, which can be arranged in
5! ways. Now, N and E can also exchange their places in two ways.
∴ Required number of words = 5! × 2 = 240
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146 Permutations and Combinations Solutions of Assignment (Level-I) (Set-1)

39. In how many ways can 9 examination papers be arranged so that the best and the worst papers are never together?
Sol. Total number of papers = 9
∴ Number of arrangements = 9!
Let best and worst paper as a single paper, then total number of papers is 8. Best and worst can be put
together in 2! ways.
∴ Number of arrangement of 8 papers in which best and worst put together = 8! × 2!
Hence, Number of arrangements of best and worst should never be together = Arrangements of total paper –
Arrangements of papers in which best and worst are together.
= 9! – 8! × 2!
= 8! (9 – 2)
= 7 × 8!
= 282240

‰ ‰ ‰

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 Level-I
Chapter 10

Permutations and Combinations

Solutions (Set-2)

[Permutations]
1. If nCx = 56 and nPx = 336, then n is equal to
(1) 7 (2) 8 (3) 6 (4) 10
Sol. Answer (2)
nP = x! nCx
x

 x! × 56 = 336

336
 x! = =6
56

 x! = 1 × 2 × 3
 x! = 3!
∴ x=3
Now, nC3 = 56

n!
 = 56
3!(n − 3)!

n(n − 1)(n − 2)(n − 3)!

 = 3! × 56
(n − 3)!

 n(n – 1) (n – 2) = 6 × 7 × 8
∴ n=8
2. How many numbers greater than 50000 can be formed with the digits 4, 5, 6, 7 and 8 if no digit being repeated?
(1) 96 (2) 256 (3) 218 (4) 126
Sol. Answer (1)
Ten thousands place cannot assume 4 since required numbers are greater than 50000.
∴ Required arrangements = 4 × 4P4 = 4 × 24 = 96
3. In how many ways can 10 soldiers stand in two rows having 5 soldiers in each row?
(1) 3628800 (2) 7257600 (3) 35400 (4) 985000

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148 Permutations and Combinations Solutions of Assignment (Level-I) (Set-2)

Sol. Answer (1)

Any 5 soldiers can stand in one row and the remaining 5 soldiers can stand in other row.
∴ Required numbers of arrangements = 10P
5 × 5P5 = 3628800
4. If the maximum number of trials required to open all locks when there are n locks and n keys is 105,
then n =
(1) 13 (2) 14 (3) 35 (4) 20
Sol. Answer (2)
To open the first lock, maximum number of trial is n.
Similarly to open subsequent locks, maximum number of trials is n – 1, n – 2, ……..

n(n + 1)
Maximum number of trials = n + (n − 1) + (n − 2) + ...1 = = 105
2

n(n + 1) = 14 × 15

∴ n = 14.
5. How many arrangements can be made out of the letters of the word COMMITTEE, taken all at a time, such
that the four vowels do not come together?
(1) 12600 (2) 45600 (3) 43200 (4) 216
Sol. Answer (3)
Number of letter in the word COMMITTEE is 9 out of these M occurs twice, T occurs twice, E occurs twice
and remaining are different.

9!
∴ Number of arrangement = 2! 2! 2! = 45360

Now, there are 4 vowels O, I, E, E in the given word, taking them as one letter, we have to arrange 6 letters
which include 2M’s and 2T’s.

6!
∴ Number of arrangements = 2! 2! = 180 ways.

4!
Number of arrangements of 4 vowels = = 12
2!
∴ Number of arrangements in which four vowels always come together = 180 × 12 = 2160.
Hence, required number of arrangements = 45360 – 2160
= 43200
6. How many different words can be made with the letters of the word NAINITAL such that each of the word begins
with L and end with T?
(1) 85 (2) 88 (3) 92 (4) 90
Sol. Answer (4)
When L and T are fixed as first and last letters of the word, then we have, only 6 letters to be arranged.

6!
Hence, required number of arrangements = = 90
2! 2! 2!

7. How many numbers greater than a million can be formed with the digits 5, 5, 2, 2, 1, 7, 6?
(1) 1320 (2) 1180 (3) 1000 (4) 1260
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Solutions of Assignment (Level-I) (Set-2) Permutations and Combinations 149
Sol. Answer (4)
For a number greater than million, we have to use all 7 seven digits which include 5 twice, 2 twice and
remaining are different.

7!
∴ Number of arrangements = 2! 2! = 1260

8. How many different signals can be given using any number of flags from 5 flags of different colours?
(1) 325 (2) 2030 (3) 1888 (4) 1920
Sol. Answer (1)
Number of signals using single flag is 5P1
Number of signals using two different flags is 5P2
Number of signals using three different flags is 5P3
Number of signals using four different flags is 5P4
Number of signals using five different flags is 5P5
∴ Required number different signals = 5P5 + 5P4 + 5P3 + 5P2 + 5P1
= 120 + 120 + 60 + 20 + 5
= 325
9. The number of permutations of n different objects taken k at a time, when repetitions are allowed is
(1) k n (2) nk (3) n ! (4) nP
k

Sol. Answer (2)

Number of permutations of n different objects, taken k at a time, when repetitions are allowed is given by nk.
10. Number of natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3, 4 if repetition is allowed
is
(1) 123 (2) 113 (3) 313 (4) 222
Sol. Answer (3)
Total one digit number = 4
Total two digit numbers = 4 × 4 = 16
Total three digit numbers = 4 × 4 × 4 = 64
Number of 4 digit numbers starting with 1 = 4 × 4 × 4 = 64
Number of 4 digit numbers starting with 2 = 4 × 4 × 4 = 64
Number of 4 digit numbers starting with 3 = 4 × 4 × 4 = 64
Number of 4 digit numbers starting with 41 = 4 × 4 = 16
Number of 4 digit numbers starting with 42 = 4 × 4 = 16
Number of 4 digit numbers starting with 431 = 4
Number of 4 digit numbers starting with 432 = 1
Total number of 4 digit numbers = 3 × 64 + 2 × 16 + 4 + 1
= 229
Hence, required number of natural number not exceeding 4321 is 4 + 16 + 64 + 229 = 313
11. Number of ways in which the letters of the word RAINBOW be arranged such that N and B are together is
(1) 2560 (2) 1540 (3) 540 (4) 1440

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150 Permutations and Combinations Solutions of Assignment (Level-I) (Set-2)

Sol. Answer (4)

Number of letters in the word RAINBOW = 7
Suppose N and B as single letter, then number of remaining letters is 6, which can be arranged themselves
in 6! ways. Also N and B are arranged in 2! ways.
∴ Required number of arrangements = 6! × 2! = 1440
12. Number of even numbers greater than 300 that can be formed with the digits 1, 2, 3, 4, 5 such that no digit
being repeated is
(1) 111 (2) 900 (3) 600 (4) 800
Sol. Answer (1)
A number will be even number if its unit digit is 2 or 4. Number greater than 300 may be of 3-digit numbers, 4-
digit numbers or 5-digit numbers.
Case-I : 3-digit numbers
When 2 is fixed at unit place, we have 3, 4 and 5 i.e., three digits for filling hundreds place and
remaining 3-digits for filling tens place. Similarly, when 4 is fixed at the unit place, hundred place
can be filled by 2 choices and the ten’s digit can be filled by 3 choices.
∴ Number of 3-digit numbers = 9 + 6
= 15
Case-II : 4-digit numbers
For unit digit, avaible choice is 2 and the remaining digits can be arranged in 4P3 ways.
Number of 4-digit numbers = 2 × 4P3
= 48
Case-III : 5-digit numbers
∴
Unit digit, can be arranged in 2 ways and the remaining digits can be arranged in 4P4 ways.
Number of 5-digit numbers = 2 × 4P4
= 48
Hence, total number of numbers greater than 300 is 15 + 48 + 48 = 111.
13. There are 6 books of physics, 3 of chemistry and 4 of biology. Number of ways in which these books be placed
on a shelf if the books of the same subject are to be together is
(1) 622080 (2) 888000 (3) 222000 (4) 413080
Sol. Answer (1)
Since, the same subject are to placed together
∴ Number of arrangements of placing Physics = 6! = 720
Number of arrangements of placing Chemistry = 3! = 6
Number of arrangements of placing Biology = 4! = 24
Set of 6 books of Physics, 3 of Chemistry and 4 of Biology can arranged themselves in 3! ways.
Required number of arrangements of sets of books = 720 × 6 × 24 × 3!
= 622080
14. A person has 5 shirts, 4 coats and 7 ties. Number of ways in which he can wear them is
(1) 100 (2) 140 (3) 96 (4) 124
Sol. Answer (2)
Number of arrangements = 5 × 4 × 7
= 140

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Solutions of Assignment (Level-I) (Set-2) Permutations and Combinations 151
15. Number of ways in which 15 different books can be arranged on a shelf so that two particular books shall not
be together is
(1) 14 × 15! (2) 13 × 14! (3) 14! × 15! (4) (15!)2
Sol. Answer (2)
Number of arrangements of 15 different books = 15!
Let two particular books as a set of one book, then remaining number of different books is 14, which can
arranged in 14!. Now, the set of 2 books also arranged themselves in 2! ways.
∴ Required number of arrangements in which a set of two particular books shall not be together
= 15! – 14! × 2!
= 15 × 14! – 14! × 2!
= 14! (15 – 2)
= 13 × 14!
16. How many 6 digit numbers can be formed out of the digits of the number 113226?
(1) 156 (2) 180 (3) 280 (4) 120
Sol. Answer (2)
6!
Number of arrangements = = 180
2! 2!
17. Number of three digit numbers such that at least one of the digits is 9, if repetitions allowed, is
(1) 252 (2) 648 (3) 864 (4) 468
Sol. Answer (1)
Number of 3 digit numbers = 9 × 10 × 10
= 900
Number of 3 digit numbers excluding 9 is = 8 × 9 × 9 = 648
∴ Number of 3 digit numbers such that at least one of the digits is 9
= 900 – 648 = 252
18. Number of ways in which the letters of the word TAMANNA be arranged is
(1) 120 (2) 420 (3) 840 (4) 500
Sol. Answer (2)
Number of letters in the word TAMANNA = 7. Out of these there are 3A’s and 2N’s and the remaining letters
are different.

7!
∴ Required number of arrangements = 3! 2! = 420

19. Number of three letter words that can be formed using only vowels but each only once is
(1) 100 (2) 50 (3) 60 (4) 80
Sol. Answer (3)
Number of vowels = 5
∴ Required number of words = 5P3 = 60
20. Number of different words that can be made using the letters of the word HALLUCINATION if all consonants
are together is
2
 7!  7! 14! 14
(1) 
 2! 2! 
(2) (3) (4)
2! 2! 7! 2! 2!

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152 Permutations and Combinations Solutions of Assignment (Level-I) (Set-2)

Sol. Answer (1)

Number of letters in the word HALLUCINATION = 13
Number of consonants = 7
Out of these there are 2 L’s, 2N’s and remaining are different.

7!
Number of arrangements of consonants =
2! 2!

Let set of consonants as single letter, then remaining letters in the word HALLUCINATION is 7, in which there
are 2A’s, 2I’s and the remaining are different

7!
∴ Number of arrangements = 2! 2!

Hence, required number of arrangements of the word HALLUCINATION if all consonants are together
2
7! 7!  7! 
= × =
2! 2! 2! 2!  2! 2! 

21. Number of 6 digit numbers that can be formed using the digit 2 two times and the digit 5 four times is
(1) 16 (2) 15 (3) 24 (4) 18
Sol. Answer (2)

6!
Required number of arrangements = = 15
2! 4!

22. Number of ways in which 15 billiard balls be arranged in a row if 3 are identical red, 7 are identical white and
remaining are identical black balls is

15! 15!
(1) 15! (2) (3) 7! 3! 5! (4)
7! 3! 5! 7! 5!

Sol. Answer (2)

15!
Required number of arrangements =
3! 5!7!

23. Number of 7 digit telephone numbers that can be formed from the digits 0, 1, 2, …………9, if each telephone
number begins with digit 5 is
(1) 610 (2) 106 (3) 10! (4) 10C
6

Sol. Answer (2)

Each digit can be filled by any one of the 10 digits.
∴ Required number of telephone numbers = 106
24. The total number of permutations of n different objects taken at least one but not more than r at a time, where
each object may be repeated any number of time is

n(n n − 1) nr − 1 n(n r − 1) n(n r − 1)

(1) (2) (3) (4)
n −1 n −1 n −1 r −1
Sol. Answer (3)
n.(n r − 1)
Required number = n + n2 + ……nr =
n −1

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Solutions of Assignment (Level-I) (Set-2) Permutations and Combinations 153
25. An n-digit number is a positive number with exactly n digits. Nine hundred distinct n digit numbers are to be
formed using only three digit 5, 7, 9. The smallest value of n for which this is possible is
(1) 6 (2) 7 (3) 8 (4) 9
Sol. Answer (2)

..................... n
th

3 ways 3 ways 3 ways

3n ≥ 900
⇒ n=7
26. The number of arrangements of the letter of the word PAPAYA in which the two ‘P’ do not appear adjacently is
(1) 40 (2) 60 (3) 80 (4) 36
Sol. Answer (1)
PAPAYA

6!
Total number of arrangements without restriction = = 60
3!2!

5!
Total number of arrangements when both (P) are together = = 20
3!
Total number of arrangement when no two ‘P’ are together
= 60 – 20 = 40
27. How many different nine digit numbers can be formed with the number 223355888 by rearranging its digits so
that the odd digits occupy even positions?
(1) 16 (2) 36 (3) 60 (4) 180
Sol. Answer (3)
223355888
No. of even positions = 4
No. of odd positions = 5

4! 5!
Total no. of arrangement = × = 6 × 10 = 60.
2!2! 2!3!

28. How many words are formed if the letters of the word GARDEN are arranged with the vowels in alphabetical
order?
(1) 120 (2) 240 (3) 360 (4) 480
Sol. Answer (3)
Vowels are A, E
In half of arrangements, A will come before E and in half of arrangement E will come before A.

6! 720
∴ Number of ways in which vowels will in alphabetical order = = = 360
2 2
29. 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
(1) 602 (2) 603 (3) 600 (4) 601
Sol. Answer (4)

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154 Permutations and Combinations Solutions of Assignment (Level-I) (Set-2)

Alphabetical order of word SACHIN is ACHINS

Total no. of word beginning from A, C, H, I, N is 5 × 5 × 4 × 3 × 2 × 1
= 600
SACHIN is the 1st word beginning from ‘S’.
∴ Rank of SACHIN is 601.
[Combinations]
30. In how many ways 7 members forming a committee out of 11 be selected so that 3 particular members must
be included?
(1) 81 (2) 70 (3) 95 (4) 125
Sol. Answer (2)
If 3 particular members are included, then remaining 4 members are selected out of the 8 members. It can be
done in 8C4 ways.

8!
8C = = 70
4 4! 4!

31. The number of words which can be formed out of the letters of the word PARTICLE, so that vowels occupy
the even place is
(1) 7! (2) 4C × 3C3 (3) 180 (4) 4! × 5!
4
Sol. Answer (4)
Number of letters in the word PARTICLE = 8
Number of vowels = 3
3 vowels can be arranged at 4 even places in 4P3 ways. After arranging the vowels, remaining consonants
can be arranged themselves in 5! ways.
∴ Required number of arrangements = 4P3 × 5! = 4! × 5!
32. The number of ways in which three girls and ten boys can be seated in two vans, each having numbered seats,
three in the front and four at the back is
(1) 14C (2) 14P (3) 14! (4) 103
12 13
Sol. Answer (2)
Total number of seats = 2( 4 + 3) = 14
Total number of persons = 3 + 10 = 13
∴ Number of required arrangements = 14P
13

33. A box contains 7 red, 6 white and 4 blue balls. Number of ways of selection of three red balls is
(1) 35 (2) 45 (3) 27 (4) 36
Sol. Answer (1)
Only red balls is selected
Number of total red balls = 7
Number of selected red balls = 3
∴ Required number of selection = 7C3 = 35
34. In a test paper there are 10 questions. Number of ways in which 6 questions to be answered is
(1) 105 (2) 210 (3) 310 (4) 220
Sol. Answer (2)
Required number of ways is 10C6

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Solutions of Assignment (Level-I) (Set-2) Permutations and Combinations 155
35. In a meeting everyone had shaken hands with everyone else, it was found that 66 handshakes were exchanged.
Number of persons present in the meeting is
(1) 17 (2) 12 (3) 13 (4) 18
Sol. Answer (2)
One handshake is performed if two persons handshake to each other. If number of persons be n, then number
of handshakes is nC2.
Now, nC2 = 66
 n(n – 1) = 132
 n(n – 1) = 12 × 11
∴ n = 12.
Hence, number of persons is 12
36. Number of all permutations of the set of four letters E, O, S, P taken two at a time is
(1) 4C × 2! (2) 4C (3) 2! (4) 4!
2 2

Sol. Answer (1)

Required number of permutations = 4P2
= 4C2 × 2! [∵ nP
r = r! nCr]
37. In a badminton tournament each player played one game with all the other players. Number of players
participated in the tournament if they played 105 games is
(1) 35 (2) 15 (3) 12 (4) 10
Sol. Answer (2)
Let number of players = n
∴ Number of games played = nC2
According to the question,
nC = 105
2

 n(n – 1) = 210
 n(n – 1) = 14 × 15
 n = 15.
38. Number of ways in which 5 plus (+) signs and 5 minus (–) signs be arranged in a row so that no two minus
signs are together is
(1) 6 (2) 7 (3) 8 (4) 10
Sol. Answer (1)
Fix any one the 5 plus (+) signs or 5 minus (–) signs as shown in the figure. Let it be + signs. Since all are
identical, therefore they can be arranged in a single way.

+ + + + +

6!
After arranging the plus (+) signs, minus (–) signs can be arranged in 6 blank box in ways
5!
[∵ Minus signs are also identical]
=6
∴ Total number of arrangements = 6 × 1 = 6.

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156 Permutations and Combinations Solutions of Assignment (Level-I) (Set-2)

39. Find the number of ways in which a person can buy 6 chocolates, if there are three types of chocolates
available. (Chocolates of same type are identical)
(1) 10 (2) 28 (3) 36 (4) 63
Sol. Answer (3)
Each chocolate can be selected in 3 ways. i.e., 3 × 3 × 3 × 3 × 3 × 3 = 36 ways
40. nC n + 2.nCr =
r+1 + Cr–1
(1) n+2Cr+1 (2) n+1C
r (3) n+1C
r+1 (4) n+2C
r
Sol. Answer (1)
n
Cr +1 + nCr −1 + 2 nCr

n
Cr + nCr +1 + nCr −1 + nCr = n +1
Cr +1 + n +1Cr = n + 2Cr +1
41. On the eve of new year each student of class XI sends cards to his classmates. If there are 40 students in
the class number of cards exchanged during the process is
(1) 1200 (2) 600 (3) 780 (4) 1560
Sol. Answer (4)
For each selection of two students number of cards exchanged is 2.
40
∴ Total number of cards = C2 × 2 = 40 × 39 = 1560
42. A box contains 3 red, 4 white and 2 black balls. The number of ways in which 3 balls can be drawn from the
box, so that at least one ball is red, is (all balls are different)
(1) 45 (2) 64 (3) 84 (4) 85
Sol. Answer (2)

Number of ways of selecting 3 balls without any restriction = 9C3 .

Number of ways when no red ball is selected = 6C3 .

Required number of ways = 9C3 − 6C3 = 64 .

43. In how many ways 6 students and 4 teachers be arranged in a row so that no two teachers are together?
(1) 604800 (2) 246800 (3) 258600 (4) 55500
Sol. Answer (1)
First of all the 6 students can be arranged in 6! ways then 4 teachers can be arranged in 7 places in 7P4 ways.
∴ Required number of arrangements = 6! × 7P4
= 720 × 840 = 604800
44. Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women
choose the chairs from amongst the chairs marked 1 to 4; and then the men select the chairs from amongst
the remaining. The number of possible arrangements is
(1) 6C × 4C2 (2) 4P × 6P3 (3) 4C + 4C3 (4) 4P + 4P3
3 2 2 2
Sol. Answer (2)

Women will occupy 2 chairs from first 4 chairs in 4C2 2! ways

Now men will occupy 3 chairs in 6C3 3!

Required number of ways = 6 C2 2! × 6C3 × 3! = 4P2 × 6P3

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Solutions of Assignment (Level-I) (Set-2) Permutations and Combinations 157
45. Harsha invites 13 guests to a dinner and places 8 of them at one table and remaining 5 at the other, the table
being round. The number of ways he can arrange the guests is

11! 13! 12!

(1) (2) 9! (3) (4)
40 40 40

Sol. Answer (3)

13
For first table, 8 guest can be selected and arranged in C8 × 7! ways.

Now remaining 5 on another table can be arranged in 4! ways

13! 13!
∴ Required number of ways = 13
C8 × 7! × 4! = 8!5! × 7! × 4! = 40

[Geometric Problems]
46. Number of different straight lines that can be formed by joining 12 different points on a plane of which 4 are
collinear is
(1) 16 (2) 61 (3) 65 (4) 37
Sol. Answer (2)
Total number of lines formed by 12 points = 12C
2

Number of lines formed by 4 points = 4C2

∴ Required number of lines = Total lines formed by 12 points – number of lines formed by 4 collinear
points + 1
= 12C – 4C2 + 1 = 61
2

47. A polygon has 90 diagonals. Number of its sides is

(1) 25 (2) 17 (3) 15 (4) 14
Sol. Answer (3)
Number of diagonals is n-sided polygon = Number of total lines – Number of sides
= nC2 – n
According to the question,
 nC
2 – n = 90

n(n − 1)
 − n = 90
2

 n = 15
48. Number of triangles that can be formed by joining the 10 non-collinear points on a plane is
(1) 136 (2) 140 (3) 128 (4) 120
Sol. Answer (4)
A triangle can be formed by joining 3 non-collinear points. Hence number of triangles is equal to the number of
ways in which 3 points can be selected out of 10 non-collinear points.
∴ Number of triangles = 10C
3 = 120
49. Let Tn denotes the number of triangles which can be formed by using the vertices of a regular polygon of n
sides. If Tn+1 – Tn = 21, then n is equal to
(1) 5 (2) 7 (3) 6 (4) 4

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158 Permutations and Combinations Solutions of Assignment (Level-I) (Set-2)

Sol. Answer (2)

Tn +1 − Tn = 21

n +1
C3 − nC3 = 21
⇒ n = 7.
50. There are m points on the line AB and n points on the line AC, excluding the point A. Triangles are formed
joining these points
(i) When point A is not included,
(ii) When point A is included.
The ratio of the number of such triangles is

m+n+2 m+n−2 m+n−2 m+n+2

(1) (2) (3) (4)
2 2 m+n m+n
Sol. Answer (3)
When A is not included

m mn(m + n − 2)
Number of triangles = C2 ⋅ nC1 + mC1 ⋅ nC2 =
2
When A is included

mn(m + n − 2) m mn(m + n )
Number of triangles = + C1 . nC1 =
2 2

m+n−2
Required ratio =
m+n
51. The number of rectangles, of any size excluding squares from the rectangle of size 8 × 7 is
(1) 784 (2) 840 (3) 896 (4) None of these
Sol. Answer (2)

7
Required number = C2 × C2 −  (9 − r )(8 − r )
9 8

r =1

 7(7 + 1)(15)  17(7)(7 + 1)

= 36 × 28 − 72 × 7 −   +
6 2

= 1008 – 504 – 140 + 476 = 840

[Miscellaneous]
52. There are four balls of different colours and four boxes of colours, same as those of balls. The number of ways
in which the balls, one each in one box, could be placed such that any ball does not go to the box of its
own colour, is
(1) 9 (2) 13 (3) 8 (4) 16
Sol. Answer (1)

 1 1 1 1
Required number of ways = 4!  1 − + − +  = 12 – 4 + 1 = 9.
1! 2! 3! 4!

53. Ramya gives a dinner party for 5 guests. The number of ways in which they may be selected from among 9
friends if two of the friends will not attend the party together is
(1) 84 (2) 133 (3) 91 (4) 126
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Solutions of Assignment (Level-I) (Set-2) Permutations and Combinations 159
Sol. Answer (3)

Total number of ways = 9C5 − 7C3 = 91

( 7C3 = Number of ways when two particular friends appear together).

54. 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 candidate, then the number of ways
in which he can vote is
(1) 6210 (2) 385 (3) 1110 (4) 5040
Sol. Answer (2)
A vote can voter maximum 4 votes and minimum 1 vote
10
∴ Required number of ways = C1 + 10C2 + 10C3 + 10C4

= 10 + 45 + 120 + 210 = 385

55. The number of ways in which a composite number N = 22325272 can be resolved into two factors which are
prime to each other is
(1) 2 (2) 22 (3) 23 (4) 24
Sol. Answer (3)

N = 22.32.52.72
Prime divisors = 4 (2, 3, 5, 7)

∴ Required number = 24−1 = 23

56. The sum of all five digit numbers than can be formed using the digits 1, 2, 3, 4, 5 when repetition of digits is
not allowed is
(1) 366000 (2) 660000 (3) 360000 (4) 3999960
Sol. Answer (4)
5 will occur at unit place = 4 × 3 × 2 = 24 times.
Similarly, 1, 2, 3, 4, 5 will occur at each place 24 times.

4 3 2 1
∴ Required sum = (1 + 2 + 3 + 4 + 5) × 24 × 104
+ (1 + 2 + 3 + 4 + 5) × 24 × 103 + (1 + 2 + 3 + 4 + 5) × 24 × 102 + (1 + 2 + 3 + 4 + 5) × 24 × 101 +
(1 + 2 + 3 + 4 + 5) × 24 × 100
= 24 × 15 (1 1 1 1 1)
= 3999960
57. The number of ways of selecting 8 books from a library which has 9 identical books each on Mathematics,
Physics, Chemistry and English is
(1) 165 (2) 166 (3) 167 (4) 168
Sol. Answer (1)

Required number of ways = coeff. of x8 in (1+ x + x 2 + ... + x 9 )4

= 8+4–1C = 11C = 165

4–1 3

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160 Permutations and Combinations Solutions of Assignment (Level-I) (Set-2)

58. How many numbers greater than 1000 but not greater than 4000 can be formed from the digits 0, 1, 2, 3, 4,
repetition of the digits being allowed?
(1) 374 (2) 375 (3) 376 (4) 377
Sol. Answer (2)

Required number of ways = 3 × 53 − 1 + 1 = 375 (One is subtracted to exclude 1000 and one is added to include
4000).

1,2,3

3 ways 5 ways 5 says 5 ways

59. The number of ways of distributing 12 different objects among three persons such that one particular person is
always getting 6 objects and the remaining two persons are getting three objects each, is

12 ! 12! 12 ! (12)!
(1) 2 (2) 2 (3) (4)
6 ! (3 ! ) 2 ! 6 ! (3 ! ) (6 ! ) 2
(3 ! ) 2 ( 6 ! ) 2

Sol. Answer (2)

12!
12
Required number of ways = C6 × 6C3 × 3C3 = 6!(3!)2

60. The total number of ways in which four boys and four girls can be seated around a round table, so that no
two girls sit together is equal to
(1) 7 ! (2) 3 ! 4 ! (3) 4 ! (4) 4 ! 4 !
Sol. Answer (2)
The number of ways in which all boys can be sit around a round table is 3! and hence they create and such
voids, and they can be arranged in 4! ways = 3! 4!.
61. There are 15 couples taking part in a single tournament, the number of ways in which they can be paired such
that no two real lives couple play in the same team is

 1 1 1 1 
(1) 15 ! 1– + – ........ –  (2) 15 !
 1! 2 ! 3 ! 15 ! 

(3) 30C (4) 30 C – 15C

2 2 1

Sol. Answer (1)

 1 1 1 1 1 
15!  1 − + − + ..... − 
 1! 2! 3! 4! 15! 

Clearly this is equal to the number of dearrangement.

62. The number of different ways of assigning 10 marks to 3 questions is
(1) 72 (2) 71 (3) 36 (4) 84
Sol. Answer (3)
Any question cannot be of less than 1 mark
Clearly the number of different ways
= 10 – 1C
3–1

9 9×8
= C2 = ⇒ 36
2 ×1

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Solutions of Assignment (Level-I) (Set-2) Permutations and Combinations 161
63. A committee of 12 is to be formed from 9 women and 8 men. Then the number of ways of selecting committees
so that men are in majority is
(1) 3202 (2) 1134 (3) 2702 (4) 1314
Sol. Answer (2)
For women to be in majority following selection can be done.
No. of women No of men No of ways
9
9 3 C9 × 8C3

9
8 4 C8 × 8C4

a
7 5 C7 × 8C5

2702

For men to be in majority.

No. of women No. of men No. of ways
9
4 8 C4 × 8C8

9
5 7 C5 × 8C7

1134

‰ ‰ ‰

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