Chapter-6

PERMUTATION & COMBINATION

PRACTICE SHEET
1. How many 3digit numbers, each less than 600, can be 12. How many words, with or without meaning can be formed by
formed from {1, 2, 3, 4, 7, 9} if repetition of digits is using all the letters of the word ‘MACHINE’ so that the
allowed? vowels occurs only the odd positions?
(a) 216 (b) 180 (a) 1440 (b) 720
(c) 144 (d) 120 (c) 640 (d) 576
2. There are four chairs with two chairs in each row. In how 13. From 7 men and 4 women a committee of 6 is to be formed
many ways can four persons be seated on the chairs, so that such that the committee contains at least two women. What is
no chair remains unoccupied? the number of ways to do this?
(a) 6 (b) 12 (a) 210 (b) 371
(c) 24 (d) 48 (c) 462 (d) 5544
3. In how many ways can the letters of the world 14. If P(32,6) = k C (32, 6), then what is the value of k?
CORPORATION be arranged so that vowels always occupy (a) 6 (b) 32
even places? (c) 120 (d) 720
(a) 120 (b) 2700
(c) 720 (d) 7200 15. What is the smallest natural number n such that n! is divisible
by 990?
4. In fall permutations of the letters of the ‘LAGAN’ are (a) 9 (b) 11
arranged as in dictionary, then what is the rank of ‘NAAGL’? (c) 33 (d) 99
(a) 48th Word (b) 49th Word
(c) 50th Word (d) 51st Word 16. What is the value of r, if P (5, r) = P (6, r 1)?
(a) 9 (b) 5
5. If a secretary and a joint secretary are to be selected from a (c) 4 (d) 2
committee of 11 members, then in how many ways can they
be selected? 17. What is the number of words formed from the letters of the
(a) 110 (b) 55 word ‘JOKE’ so that the vowels and consonants alternate?
(c) 22 (d) 11 (a) 4 (b) 8
(c) 12 (d) None of these
6. In how many ways can 7 persons stand in the form of a ring?
(a) P(7, 2) (b) 7! 18. If C (n, 12) = C (n, 8), then what is the value of C(22, n)?
(c) 6! (d) 7!/2 (a) 131 (b) 231
(c) 256 (d) 292
7. On a railway route there are 20 stations. What is the number 19. In a football championship 153 matches were played. Every
of different tickets required in order that it may be possible to team played one match with each other team. How many
travel from every station to every other station? teams participated in the championship?
(a) 40 (b) 380 (a) 21 (b) 18
(c) 400 (d) 420 (c) 17 (d) 15
8. What is the number of five  digit numbers formed with 0, 1, 20. How many times does the digit 3 appear while writing the
2, 3, 4 without any repetition of digits? integers from1 to 1000?
(a) 24 (b) 48 (a) 269 (b) 308
(c) 96 (d) 120 (c) 300 (d) None of these
9. A group consists of 5 men and 5 women. If the number of 21. In how many ways can a committee consisting of 3 men and 2
different fiveperson committees containing k men and (5k) women be formed from 7 men and 5 women?
women is 100, what is the value of k?
(a) 2 only (b) 3 only (a) 45 (b) 350
(c) 2 or 3 (d) 4 (c) 700 (d) 4200
10. If 7 points out of 12 are in the same straight line, then what is 22. What is the number of signals that can be sent by 6 flags of
the number of triangles formed? different colours taking one or more at a time?
(a) 84 (b) 175
(c) 185 (d) 201 (a) 21 (b) 6
(c) 720 (d) 1956
11. In how many ways can 3 books on Hindi and 3 books on
English be arranged in a row on a shelf, so that not all the 23. What is the number of words that can be formed from the
Hindi Books are together? letters of the word ‘UNIVERSAL’, the vowels remaining
(a) 144 (b) 360 always together?
(c) 576 (d) 720 (a) 720 (b) 1440
(c) 17280 (d) 21540

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24. A team of 8 players is to be chosen from a group of 12 (a) 60 (b) 108
players. Out of the eight players one is to be elected as captain (c) 120 (d) 216
and another vice-captain. In how many ways can this be
done? 26. What is the number of ways of arranging the letters of the
(a) 27720 (b) 13860 word ‘BANANA’ so that no two N’s appear together?
(c) 6930 (d) 495 (a) 40 (b) 60
(c) 80 (d) 100
25. What is the number of three-digit odd numbers formed by
using the digits 1, 2, 3, 4, 5, 6 if repetition of digits is
allowed? :

ANSWER KEYS

1. c 2. c 3. d 4. b 5. a 6. c 7. b 8. c 9. c 10. c
11. c 12. d 13. b 14. d 15. b 16. c 17. b 18. b 19. b 20. c
21. b 22. d 23. c 24. a 25. b 26. a

Solutions
Sol.1. (c) 4! Sol.10. (c)
There digit number less than 600 will have there are = 12 words. Next the 37th Number of triangles formed from 12 point
2! 12
first element 100, and last element 599. = C3
word starts with L that comes next in
First place will not have digit more than 6, 7
hence, 7 and 9 cannot be taken: So first
dictionary order there are 12 words starting Since 7 pars are collinear, then C3
with L. This accounts up to the 48 words. triangle will not be formed so.
digit can be selected in 4 ways. Second
The 49th word is ‘NAAGL’ =12C3–7C3
digit can be selected in 6 ways and since
repetitions of digits are allowed, third digit Sol.5. (a)
can also be selected in 6 ways: So, number Selection of 2 members out of 11 has 11 C2 12! 7! 12.11.10 7.6.5
of ways are 4 × 6 × 6 = 144. =   
number of ways of arrangement = 55 × 2 = 3!9! 3!4! 3.2.1 3.2.1
Sol.2. (c) 110
First chair can be occupied in 4 ways and Sol.6. (c) =220–35=185
second chair can be occupied in 3 ways, Number of ways in which 7 person can
third chair can be occupied in 2 ways and stand in the form of ring = (7 1)! = 6! Sol.11. (c)
last chair can be occupied in one ways only. Sol.7. (b) Total number of arrangement = 6! = 720
So total number of ways = 4 × 3 × 2 × 1 = From each railway station, there are 19 Total number of arrangement while all the
24 different tickets to be issued. There are 20 Hindi books are together = 4!×3!= 24 × 6 =
Sol.3. (d) railway station. 144
CORPORATION is 11 letter word. So, total number of tickets = 20× 19 = 380  The number of ways, in which books are
It has 5 vowels (O, O, O, A, I) and 6 Sol.8. (c) arranged, while all the Hindi books are not
consonants (C, R, P, R, T, N) To make a 5 digit number, 0 cannot come together
In 11 letters, there are 5 even places (2nd, in the beginning. So, it can be filled in 4 = 720  144 = 576
4th, 6th, 8th and 10th positions) ways. Rest of the places can be filled in 4!
Sol.12. (d)
5! Ways. So total number of digit formed = 4 There are three vowels and they have four
5 vowels can take 5 even places in
3! × 4! = 4 × 24 = 96 odd places to arrange. Other letters are four
ways Sol.9. (c) and has four places to arrange.
(∵ Since O is repeated thrice) K men selected out of 5 and 5  k women  The number of words = 4 P3  4!
Similarly, 6 consonants can take 6 odd out of 5. These are 5 C k and 5 C5  k
4!
6! According to problem:   4!  576
places in ways (4  3)!
1C  C5–1C = 100
5C 5
2! Sol.13. (b)
(∵ R is repeated twice) The required number of ways
5! 6! 5! 5!
= 20  k!5  k !  5  k !5!  100 =11C6–(7C6  4C0 + 7C5  4C1)
 Total number of ways = 
3! 2!
× 360 11 10  9  8  7  76 
2 =  7   4
= 7200  5  5 4  3 2  2 
    100
Sol.4. (b)  k!5  k ! 
Starting with the letter A and arranging the =462 – (7 + 84) = 371
other four letters, there are 24 words. There
5!
are the first 24 words. Then starting with G   10
that comes next in dictionary order and k!5  k ! Sol.14. (d)
33 32
arranging A, A, L, N in different ways, This is true of k = 2 or 3. Since P6  k C6

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 32! 32! 1 Consider the word UNIVERSAL
  k. 
(32  6)! 6!(32  6)! n  8n  9n  10n  11n  12!8! Total no. of vowels= U,I,E,A=4
 k  6!  720
Let us consider these as a single letter

UNIVERSAL
Sol.15. (b) 1 1
 Then, total No. of letters =6
Consider option ‘a’ 12 1110  9 n  8n  9n  10n  11 Then, number of ways to arrange
Let us take n = 9 (n–8) (n–9) (n-10) (n–11) them=6!=720.
Since, 9! = 9×8×7×6×5×4×3×2×1=362880 =12  11  10  9 But vowels can also arranged in 4! Or 24
Which is not divisible by 990. n–8=12, n–9=11, n–10=10 and n–11=9 ways.
Now assume, n = 11 n=20 Hence, total number of ways=720 X 24
Since 11! = 39916800
C(22, n) = 22C20 =17280
Which is divisible by 990. Sol.24. (a)
22! 22  21
Thus, required smallest natural number 11 =   231 Total no. of players=12
2!20! 2
Sol.16. (c) No. of chose players=8
Given P (5, r) = P (6, r 1) Sol.19. (b)
Number of ways to choose 8 players from
5Pr = 6Pr–1 Let total no. of team participated in a
12
championship be n. Since every team
players
played one match with each other team.
5! 6! 12! 12 1110  9  8!
  n! 12 C8    495
5  r ! 6  r  1 ! nC2=153 
2! n  2!
 153 8!4! 8!4!
Since, out of the 8 players 1 is to be elected
n n  1n  2! n n  1 as
5! 6!   153   153
  2! n  2!
5  r ! 7  r  ! 2 captain and another vice-captain
n(n–1)=306 therefore number of ways to choose a
captain
5! 6  5! n2 – n – 306 = 0
  and a vice-captain.
7  r ! 7  r 6  r 5  r  ! n(n–18) + 17 (n – 18) = 0
8 C1 7C1  8  7  56  n
C1  n 
n = 18, – 17
(7–r)(6–r) = 6 N cannot be negative Here, required number of ways = 495×56 =
27720
n  –17
Sol.25. (b)
42–13r+r2 = 6  n =18 Total No. of digits=6
Sol.20. (c) To form a odd numbers we have only 3
r2 – 13r + 36 = 0 Before 1000 there are one digit, two digits choice
and three digits numbers. for the unit digits.
r2 – 9r – 4r + 36 = 0 Number of ties 3 appear in one digit Now, Extreme left place can be filled in 6
number = 20 × 9 ways
(r–9) (r–4)=0 Number of times 3 appear in two digit the middle.
number = 11 × 9
Required number of numbers = 6 x 6 x 3
Number of times 3 appear in three digit
r = 4 (r9) numbers = 21
= 108
Sol.17. (b) Hence total number of times the digit 3
Sol.26. (a)
Total number of letters = 4 Total no. of letters in BANANA = 6
appear while writing the integers from 1 to
No. of vowels = 2 No. of repeated letter N = 2
1000
No. of consonants = 2 No. of repeated letter A =2
= 180 + 99 + 21= 300
Possibilities of words formed from the Therefore , Number of ways that can be
Sol.21. (b)
letters of word “JOKE” are Total no. of Men=7
formed
JOKE, KOJE, KEJO, JEKO, EJOK, EKOJ, by using the words
Total no. of women=5
OKEJ, OJEK Required number of ways= 7C3X5C2 ‘BANANA’ = 6!  6  5  4  3!  60
Thus, required number of words = 8 3!2! 3! 2!
7! 5! 7  6  5 5  4
Sol.18. (b)     Number of ways in which two N comes
Given C (n, 12) = C (n, 8) 3!4! 2!n ! 3 2 2 together= 5! =20
nC12 = nC8 7  5 10  35 10  350 3!
n! n! Sol.22. (d)  Required number of ways
  = 60 20= 40
n  12!12! n  8!8! Required number of ways
1 6 P16P2 6P3 6P4 6P5 6P6

n  12! 12 1110  9  8! = 6 + 30 + 120 + 360 + 720 + 720 =1956
Sol.23. (c)

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 NDA PYQ
1. A, B, C, D and E are coplanar points and three of them lie in (a) 540 (b) 1260
a straight line. What is the maximum number of triangles (c) 3780 (d) 5040
that can be drawn with these points as their vertices? [NDA (II) - 2012]
(a) 5 (b) 9 12. What is the number of diagonals which can be drawn by
(c) 10 (d) 12 joining the angular points of a polygon of 100 sides?
[NDA (I) - 2011] (a) 4850 (b) 4950
2. Using the digits 1,2, 3,4 and 5 only once, how many (c) 5000 (d) 10000
numbers greater than 41000 can be formed? [NDA-2012(2)]
(a) 41 (b) 48 13. In how many ways can the letters of the word ‘GLOOMY’
(c) 50 (d) 55 be arranged so that the two O’s should not be together?
[NDA (I) - 2011]
3. What is the value of n, if P(15,n – 1): P(16, n – 2) = 3 : 4 ? (a) 240 (b) 480
(a) 10 (b) 12 (c) 60 (d) 720
(c) 14 (d) 15 [NDA (I) - 2013]
[NDA (I) - 2011] 14. If P (77, 31) =x and C (77, 31) = y, then which one of the
4. In how many ways 6 girls can be seated in two chairs? following is correct?
(a) 10 (b) 15 (a) x = y (b) 2x = y
(c) 24 (d) 30 (c) 77x = 31y (d) x > y
[NDA (I) - 2011] [NDA (I) - 2013]
5. 5 books are to be chosen from a lot of 10 books. If m is the 15. A bag contains balls of two colours, 3 black and 3 white.
number of ways of choice when one specified book is What is the smallest number of balls which must be drawn
always included and n is the number of ways of choice when from the bag, without looking, so that among these there are
a specified book is always excluded, then which one of the two of the same colour?
following is correct? (a) 2 (b) 3
(a) m > n (b) m = n (c) 4 (d) 5
(c) m = n - 1 (d) m= n - 2 [NDA-2013(1)]
[NDA (I) - 2011] n
6. What is the total number of combination of n different things 16. What is  Cn, r  equal to?
r 0
taken 1, 2, 3,….,n at a time?
(a) 2n+1 (b) 22n+1 (a) 2n –1 (b) n
(c) 2 n-1 (d) 2n −1 (c) n! (d) 2n
[NDA (I) - 2011] [NDA-2013(2)]
17. If C(28,2r) = C(28, 2r –4), then what is r equal to?
7. What is the value of 
 ?
n P n, r
(a) 7 (b) 8
r 1 r! (c) 12 (d) 16
(a) 2n-1 (b) 2 n [NDA-2013(2)]
(c) 2n-1 (d) 2 n + 1 18. How many different words can be formed by taking four
[NDA (II) - 2011] letters out of the letters of the word 'AGAIN' if each word
8. There are 4 candidates for the post of a lecturer in has to start with A ?
Mathematics and one is to be selected by votes of 5 men. (a) 6 (b) 12
What is the number of ways in which the votes can be (c) 24 (d) None
given? [NDA (I) - 2014]
(a)1048 (b) 1072 19. Out of 7 consonants and 4 vowels, words are to be formed
(c)1024 (d) 625 by involving 3 consonants and 2 vowels. The number of
[NDA (II) - 2011] such words formed is :
9. How many diagonals will be there in an n-sided regular (a) 25200 (b) 22500
polygon? (c) 10080 (d) 5040
n n  1 n n  3 [NDA (I) - 2014]
(a) (b) For the next three (03) items that follow:
2 2
n n  1 Given that C(n,r) : C(n, r+1) = 1:2 and C(n, r+1) : C(n, r+2)
(c) n2–n (d) = 2:3.
2
20. What is n equal to?
[NDA (II)-2011]
(a) 11 (b) 12
10. What is the number of ways that 4 boys and 3 girls can be
(c) 13 (d) 14
seated so that boys and girls alternate?
[NDA-2014(1)]
(a) 12 (b) 72
21. What is r equal to?
(c) 120 (d) 144
(a) 2 (b) 3
[NDA (I) - 2012]
(c) 4 (d) 5
11. The number of permutations that can be formed from all the
[NDA-2014(1)]
letters of the word ‘BASEBALL’ is:
22. What is P(n,r) : C(n,r) equal to?

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 (a) 6 (b) 24 [NDA (I) - 2016]
(c) 120 (d) 720 34. What is the number of different messages that can be
[NDA-2014(1)] represented by three 0’s and two 1’s?
23. What is the number of ways in which one can post 5 letters (a) 10 (b) 9
in 7 letters boxes? (c) 8 (d) 7
(a) 75 (b) 35 [NDA (I) - 2016]
7
(c) 5 (d) 2520 35. What is the number of odd integers between 1000 and 9999
[NDA (II) - 2014] with no digit repeated?
24. What is the number of ways that a cricket team of 11 players (a) 2100 (b) 2120
can be made out of 15 players? (c) 2240 (d) 3331
(a) 364 (b) 1001 [NDA (II) - 2016]
(c) 1365 (d) 32760 36. A five-digit number divisible by 3 is to be formed using the
[NDA (II) - 2014] digits 0, 1, 2, 3 and 4 without repetition of digits. What is the
25. How many words can be formed using all the letters of the number of ways this can be done?
word ‘NATION’ so that all the three vowels should never (a) 96
come together? (b) 48
(a) 354 (b) 348 (c) 32
(c) 288 (d) None of these (d) No number can be formed
[NDA (I) - 2015] [NDA (II) - 2016]
1

37. Out of 15 points in a plane, n points are in the same straight
26. What is  Cn equal to?
n r
line. 445 triangles can be formed by joining these points.
r 0
What is the value of n?
(a) n+1C1 (b) n+2Cn (a) 3 (b) 4
(c) n+3Cn (d) n+2Cn+1 (c) 5 (d) 6
[NDA-2015(1)] [NDA (II) - 2016]
27. The number of ways in which a cricket team of 11 players be 5
chosen out of a batch of 15 players so that the captain of the
team is always included, is
38. What is 47C4 + 51C3 + 
j 2
52  j
C3 equal to?

(a) 165 (b) 364 (a) 52C4 (b) 51C5

(c) 1001 (d) 1365 (c) 53C4 (d) 52C5
[NDA (I) - 2015] [NDA (II) - 2016]
28. If different words are formed with all the letters of the word 39. The number of different words (eight-letter words) ending
‘AGAIN’ and are arranged alphabetically among themselves and beginning with a consonant which can be made out of
as in a dictionary, the word at the 50th place will be the letters of the word 'EQUATION' is
(a) NAAGI (b) NAAIG (a) 5200 (b) 4320
(c) IAAGN (d) IAANG (c) 3000 (d) 2160
[NDA (I) - 2015] [NDA (I) - 2017]
29. The number of 3-digit even numbers that can be formed 40. Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Then the number of
from the digits 0, 1, 2, 3, 4 and 5, repetition of digits being subsets of A containing two or three elements is:
not allowed, is (a) 45 (b) 120
(a) 60 (b) 56 (c) 165 (d) 330
(c) 52 (d) 48 [NDA-2017(1)]
[NDA (II) - 2015] 41. Three-digit numbers are formed from the digits 1, 2 and 3 in
30. The number of ways in which 3 holiday tickets can be given such a way that the digits are not repeated. What is the sum
to 20 employees of an organization if each employee is of such three-digit numbers?
eligible for any one or more of the tickets, is (a) 1233 (b) 1322
(c) 1323 (d) 1332
(a) 1140 (b) 3420 [NDA-2017(1)]
(c) 6840 (d) 8000 42. The value of [C(7,0) + C(7,1)] + [C(7,1) + C(7,2)]
NDA (II) - 2015] +……………..+[C(7,6)+C(7,7)] is
31. A polygon has 44 diagonals. The number of its sides is (a) 254 (b) 255
(c) 256 (d) 257
(a) 11 (b) 10 [NDA-2017(1)]
(c) 8 (d) 7 43. A tea party is arranged for 16 people along two sides of a
[NDA (II) - 2015] long table with eight chairs on each side. Four particular
32. What is the number of four-digit decimal numbers in which men wish to sit on one particular side and two particular men
no digit is repeated? on the other side. The number of ways they can be seated is
(a) 3024 (b) 4536 (a) 24 × 8! × 8! (b) (81)3
(c) 5040 (d) None (c) 210 × 8! × 8! (d) 16!
[NDA (I) - 2016] [NDA (II) - 2017]
33. What is the number of ways in which 3 holiday travel tickets 44. How many different permutations can be made out of the
are to be given to 10 employees of an organization, if each letters of the word 'PERMUTATION'?
employee is eligible for any one or more of the tickets? (a) 19958400 (b) 19954800
(a) 60 (b) 120 (c) 19952400 (d) 39916800
(c) 500 (d) 1000 [NDA (II) - 2017]

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45. What is the number of triangles that can be formed by (c) 12 (d) 16
choosing the vertices from a set of 12 points in a plane seven [NDA-2019(1)]
of which lie on the same straight line? 57. What is C(47,4) + C(51,3) + C(50,3) + C(49,3) + C(48,3) +
(a) 185 (b) 175 C(47,3) equal to?
(c) 115 (d) 105 (a) C( 47,4) (b) C(52,5)
[NDA (I) - 2018] (c) C(52,4) (d) C(47,5)
46. How many four-digit numbers divisible by 10 can be formed [NDA – 2019(2)]
using 1, 5, 0, 6, 7 without repetition of digits? 58. If n! has 17 zeros, then what is the value of n?
(a) 95
(a) 24 (b) 36 (b) 85
(c) 44 (d) 64 (c) 80
[NDA (I) - 2018] (d) no such value of n exists
47. How many numbers between 100 an 1000 can be formed [NDA-2019(2)]
with the digits 5, 6, 7, 8, 9, if the repetition of digits is not 59. If P(n,r) = 2520 and C(n,r) = 21, then what is the value of
allowed? C( n + 1, r + 1) ?
(a) 32 (b) 53 (a) 7 (b) 14
(c) 120 (d) 60 (c) 28 (d) 56
[NDA (I) - 2018] [NDA (II) - 2019]
48. What is C(n, r) + 2C (n, r–1) + C(n, r–2) equal to? 60. What is the number of diagonals of an octagon?
(a) C(n+1, r) (b) C(n–1, r+1) (a) 48 (b) 40
(c) C(n, r+1) (d) C(n+2, r) (c) 28 (d) 20
[NDA-2018(1)] [NDA-2019(2)]
49. Let x be the number of integers lying between 2999 and 61. If C(20, n + 2) = C(20, n − 2) , then what is n equal to ?
8001 which have at least two digits equal. Then x is equal to: (a) 18 (b) 25
(a) 2480 (b) 2481 (c) 10 (d) 11
(c) 2482 (d) 2483 [NDA 2020]
[NDA-2018(2)] 62. What is the number of ways in which the letters of the word
50. What is the sum of all three-digit numbers that can be 'ABLE' can be arranged so that the vowels occupy odd
formed using all the digits 3, 4 and 5, when repetition of places?
digits is not allowed? (a) 2 (b) 4
(a) 2664 (b) 3882 (c) 6 (d) 8
(c) 4044 (d) 4444 [NDA 2020]
[NDA (II) - 2018] 63. What is the maximum number of points of intersection of 5
51. The total number of 5-digit numbers that can be composed non-overlapping circles
of distinct digits from 0 to 9 is (a) 10 (b) 15
(a) 45360 (b) 30240 (c) 20 (d) 25
(c) 27216 (d) 15120 [NDA 2020]
[NDA (II) - 2018] 64. How many 5-digit prime numbers can be formed by using the
52. There are 17 cricket players, out of which 5 players can bowl digits 1, 2, 3, 4 5. If the repetition of digits is not allowed?
.In how many ways can a team of 11 players be selected to (a) 5 (b) 4
as to include 3 bowlers? (c) 3 (d) 0
(a) C(17,11) (b) C(12,8) [NDA (I) - 2021]
(c) C(17,5)×C(5,3) (d) C(5,3) ×C(12,8) 65. In how many ways can a team of 5 players be selected from 8
[NDA (II) - 2018] players so as not to include a particular player?
53. How many three-digit even numbers can be formed using (a) 42 (b) 35
the digits 1, 2, 3, 4 and 5 when repetition of digits is not (c) 21 (d) 20
allowed? [NDA (I) - 2021]
(a) 36 (b) 30 66. If an = n (n!), then what is a1 + a2 + a3 + …. + a10
(c) 24 (d) 12 (a) 10! – 1 (b) 11! + 1
[NDA (I) - 2019] (c) 10! + 1 (d) 11! – 1
54. From 6 programmers and 4 typists, an office wants to recruit [NDA (II) 2021]
5 people. What is the number of ways this can be done so as 67. Let S = (2, 3, 4, 5, 6, 7, 9). How many different 3-digit
to recruit at least one typist? numbers (with all digits different) from S can be made which
(a) 209 (b) 210 are less than 500?
(c) 246 (d) 242 (a) 30 (b) 49
[NDA (I) - 2019] (c) 90 (d) 147
55. There are 10 points in a plane. No three of these points are in [NDA (II) - 2021]
a straight line. What is the total number of straight lines 68. Consider the digit 3, 5, 7, 9. What is the number of 5-digit
which can be formed by joining the points? numbers formed by these digits in which each of these four
digits appears?
(a) 90 (b) 45 (a) 240 (b) 180
(c) 40 (d) 30 (c) 120 (d) 60
[NDA (I) - 2019] [NDA (II) - 2021]
56. If c(20, n+2) = C(20, n–2), then what is n equal to? 69. If C (n, 4),C(n, 5) and C(n, 6) are in AP, then what is the
(a) 8 (b) 10 value of n?

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 (a) 7 (b) 8 (a) 4 (b) 6
(c) 9 (d) 10 (c) 7 (d) 8
[NDA (II) - 2021] [NDA (I) - 2022]
70. How many 4 letter words (with or without meaning) 80. What is the value of 2 (21) + 3 (3  2  1) + 4 (4  3  2  1)
containing two vowels can be constructed using only the + 5 (5  4  3  2  1) + …. …. …. + 9 (9  8  7  6  5 
letters (without repetition) of the word ‘LUCKNOW’? 4  3  2  1) + 2?
(a) 240 (b) 200 (a) 11! (b) 10!
(c) 150 (d) 120 (c) 10 + 10! (d) 11 + 10!
[NDA (II) - 2021] [NDA (I) - 2022]
71. Suppose 20 distinct points are placed randomly on a circle. 81. How many four digit natural numbers are there such that all
Which of the following statements is/are correct? of the digits are odd?
1.The number of straight lines that can be drawn by joining (a) 625 (b) 400
any two of these points is 380. (c) 196 (d) 120
2.The number of triangles that can be drawn by joining any [NDA 2022 (II)]
three of these points is 1140. 82. If different permutations of the letters of the word
Select the correct answer using the code given below: ‘MATHEMATICS’ are listed as in a dictionary, how many
(a)1 only (b) 2 only words (with or without meaning) are there in the list before
(c) Both 1 and 2 (d) Neither 1 nor 2 the first word that starts with C?
[NDA (II) - 2021] (a) 302400 (b) 403600
72. Consider a regular polygon with 10 sides. What is the number (c) 907200 (d) 1814400
of triangles that can be formed by joining the vertices which [NDA 2022 (II)]
have no common side with any of the sides of the polygon? Consider the following for the next three (03) items that
(a) 25 (b) 50 follow:
(c) 75 (d) 100 Consider the word ‘QUESTION’:
[NDA (II) - 2021] 83. How many 4-letter words each of two vowels and two
73. If C(3n, 2n) = C(3n, 2n–7), then what is the value of C(n, n– consonants with or without meaning, can be formed?
5)? (a) 36 (b) 144
(a) 42 (b) 35 (c) 576 (d) 864
(c) 28 (d) 21 [NDA 2022 (II)]
[NDA (I) - 2022] 84. How many 8-letter words with or without meaning, can be
74. What is the value of formed such that consonants and vowels occupy alternate
C(51,21)–C(51,22)+C(51,23)–C(51,24) +C(51,25)– positions?
C(51,26)+C(51,27)–C(51,28) +C(51,29)–C(51,30)? (a) 288 (b) 576
(a) C(51,25) (b) C(51, 27) (c) 1152 (d) 2304
(c) C(51,51) – C(51,0) (d) C(51,25) – C(51,27) [NDA 2022 (II)]
[NDA (I) - 2022] 85. How many 8-letter words with or without meaning, can be
75. How many odd numbers between 300 and 400 are there in formed so that all consonants are together?
which none of the digits is repeated? (a) 5760 (b) 2880
(a) 32 (b) 36 (c) 1440 (d) 720
(c) 40 (d) 45 [NDA 2022 (II)]
[NDA (I) - 2022] 86. Consider the following statements for a fixed natural number
76. Consider the following statements: n:
n! 1.C(n, r) is greatest if n = 2r
1. is divisible by 6, where n > 3 2.C(n, r) is greatest if n = 2r – 1 and n = 2r + 1.
3!
Which of the statements given above is/are correct?
n!
2.  3 is divisible by 7, where n > 3 (a) 1 only (b) 2 only
3! (c) both 1 and 2 (d) Neither 1 nor 2
Which of the above statements is/are correct? [NDA – 2023 (1)]
(a) 1 only (b) 2 only 87. M parallel lines cut n parallel lines giving rise to 60
(c) Both 1 and 2 (d) Neither 1 nor 2 parallelograms. What is the value of (m + n)?
[NDA (I) - 2022] (a) 6 (b) 7
77. In how many ways can a term of 5 players be selected out of 9 (c) 8 (d) 9
players so as to exclude two particular players? [NDA – 2023 (1)]
(a) 14 (b) 21 88. Let x be the number of permutation of the word
(c) 35 (d) 42 ‘PERMUTATIONS’ and y be the number of permutations of
[NDA (I) - 2022] the word ‘COMBINATIONS’. Which one of the following is
78. How many permutations are there of the letters of the word correct?
‘TIGER’ in which the vowels should not occupy the even (a) x = y (b) y = 2x
positions? (c) x = 4y (d) y = 4x
(a) 72 (b) 36 [NDA – 2023 (1)]
(c) 18 (d) 12 89. 5-digit numbers are formed using the digits 0, 1, 2, 4, 5
[NDA (I) - 2022] without repetition. What is the percentage of numbers which
79. What is the maximum value of n such that 5n divides are greater than 50,000.
(30!+35!), where n is a natural number? (a) 20% (b) 25%

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 100 110 (a) 44440 (b) 46460
(c) % (d) % (c) 46440 (d) 64440
3 3
[NDA-2024 (1)]
[NDA – 2023 (1)] 99. A man has 7 relatives (4 women and 3 men). His wife also
90. Consider the following statements: has 7 relatives (3 women and 4 men). In how many wasy can
1.(25)! + 1 is divisible by 26 they invite 3 women and 3 men so that 3 of them are men’s
2.(6)! + 1 is divisible by 7 relatives and 3 of them are his wife’s relatives?
Which of the above statement is/are correct? (a) 340 (b) 484
(a) 1 only (b) 2 only (c) 485 (d) 469
(c) both 1 and 2 (d) neither 1 nor 2 [NDA-2024 (1)]
[NDA – 2023 (1)] 100. A triangle PQR is such that 3 points lie on the side PQ, 4
91. What is the number of 6-digit numbers that can be formed points on QR and 5 points on RP respectively. Triangles are
only by using 0, 1, 2, 3, 4 and 5 (each once); and divisible by constructed using these points as vertices. What is the number
6? of triangles so formed?
(a) 96 (b) 120 (a) 205 (b) 206
(c) 192 (d) 312 (c) 215 (d) 220
[NDA – 2023 (1)] [NDA-2024 (1)]
Consider the following for the next two (02) items that 101. If 26! = n8k, where k and n are positive integers, then what is
follow: the maximum value of k?
Consider the sum S = 0! + 1! + 2! + 3! + 4! + …… + 100! (a) 6 (b) 7
92. If the sum S is divided by 8, what is the remainder? (c) 8 (d) 9
(a) 0 [NDA-2024 (1)]
(b) 1 102. What is the maximum number of possible points of
(c) 2 intersection of four straight lines and a circle (intersection is
(d) cannot be determined between lines as well as circle and lines) ?
[NDA – 2023 (1)] (a) 6 (b) 10
93. If the sum S is divided by 60, what is the remainder? (c) 14 (d) 16
(a) 1 (b) 3 [NDA-2024 (2)]
(c) 17 (d) 34 103. In how many ways can the letters of the word INDIA be
[NDA – 2023 (1)] permutated such that in each combination, vowels should
94. If 1! + 3! + 5! + 7! + …..+ 199! is divided by 24, what is the occupy odd position?
remainder? (a) 3 (b) 6
(a) 3 (b) 6 (c) 9 (d) 12
(c) 7 (d) 9 [NDA-2024 (2)]
[NDA-2023 (2)] 104. The letter of the word EQUATION are arranged in such a
95. What is the maximum number of points of intersection of 10 way that all vowels as well as consonants are together. How
circles? many such arrangements are there?
(a) 45 (b) 60 (a) 240 (b) 720
(c) 90 (d) 120 (c) 1440 (d) 1620
[NDA-2023 (2)] [NDA-2024 (2)]
96. How many four-digit natural numbers are there such that all 105. In how many ways can a student choose (n – 2) courses out of
of the digits are even? n courses if 2 courses are compulsory?
(a) 625 (b) 500 (a) (n – 3) (n – 4) (b) (n – 1) (n – 2)
(c) 400 (d) 256 (c) (n – 3) (n – 4)/2 (d) (n – 2) (n – 3)/2
[NDA-2024 (1)] [NDA-2024 (2)]
97. Four digit numbers are formed by using the digits 1, 2, 3, 5 106. How many four digit numbers are there having all digits are
without repetition of digits. How many of them are divisible odd?
by 4? (a) 625 (b) 400
(a) 120 (b) 24 (c) 196 (d) 120
(c) 12 (d) 6 [NDA-2024 (2)]
[NDA-2024 (1)]
98. What is the sum of all four digit numbers formed by using all
digits 0, 1, 4, 5 without repetition of digits?

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 ANSWER KEY

1. b 2. b 3. c 4. d 5. b 6. d 7. a 8. d 9. b 10. d
11. d 12. a 13. a 14. d 15. a 16. d 17. b 18. c 19. a 20. d
21. c 22 b 23. a 24. c 25. c 26. a 27. c 28. b 29. c 30. d
31. a 32. b 33. d 34. a 35. c 36. d 37. c 38. a 39. b 40. a
41. d 42. a 43. c 44. a 45. a 46. a 47. d 48. d 49. b 50. a
51. c 52. d 53. d 54. c 55. b 56. b 57. c 58. d 59. c 60. d
61. c 62. b 63. c 64. d 65. b 66. d 67. c 68. a 69. a 70. a
71. b 72. b 73. d 74. c 75. a 76. d 77. b 78. b 79. c 80. b
81. a 82. c 83. d 84. c 85. b 86. c 87. d 88. c 89. b 90. b
91. d 92. c 93. d 94. c 95. c 96. b 97. d 98. d 99. c 100. a
101. b 102. c 103. b 104. c 105. d 106. a

Solutions
Sol.1. (b)   n  14  n  21  0 Required no. of ways = 4! x 3! = 144
Number of triangles using 5 points out of three Sol.11. (d)
are on a  n  14  n  21 There are total 8 letters in the word
Straight line = 5C3 − 5C3 = Sol.4. (d) BASEBALL, in which we have 2 B’s 2A’s and
5! Required number of ways= 6C2  2 2 L’s.
1
3!2! = 6  5 = 30  Required No. of permutations= 8!
= 10 − 1 = 9 Sol.5. (b) 2! 2! 2!
Sol.2. (b) Number of ways when one specified book is = 8  7  6  5  4  3  2 1
We have to construct 5 digit numbers which  5040
always included. So, 4 books can be chose from 8
are greater than 41000. So, we have only 2 ways
to choose 5th digit.
remaining 9 ways: Sol.12. (a)
= m = 9C4 no.of diagonals
5th 4th 3rd 2nd 1st Number of ways when one specified book is
always excluded. So, 5 books can be chose from = n(n  3)  100  97  4850
remaining 9 ways: 2 2

[ Only 4 or 5 can come at 5th place] = n = 9C5 Sol.13. (a)
Thus, for 4th place we have 4 ways to choose mn total words formed by GLOOMY – words
formed by taking both O together
digits, Sol.6. (d)
For 3rd place we have 3 ways. Since, combinations of taking 1,2,3,….things at 6!
 5!  240
For 2nd place we have 2 ways a times are nC1 , nC2 , nC3 ……..… nCn 2!
And for unit place we have only 1 way.
Required number of ways=2 x 4 x 3 x 2 x 1=48
 Total number of combinations Sol.14. (d)
As we know P(n,r)=r!C(n,r)
n C1  n C2  ......  n Cn  2n  1
Sol.3. (c)  From the question, we have
Let P (15,n-1) : P(16, n-2) = 3:4 Sol.7. (a) x = r!(y)
15 Consider n
P  n, r  n
 16
Pn 1 3
 
r 1 r!
  C  n, r 
r 1
Here r = 31
x = (31)! y
Pn 2 4
n C1  n C2  ......  n Cn  2n  1 so x > y

15!

16  n  2   3 Sol.15. (b)
Sol.8. (d)
15  n  1! 16! 4 There are 4 candidates.
if we take 2 balls together then it may be of
different colour

15!

18  n !  3 This means there are 4 blanks spaces for 1 post. if we take 3 balls together then minimum 2 balls
16  n ! 16! 4
Now, that 1 post is to be selected by votes of 5
men.
must be of same colour.
18  n !  3 So, All 4 places can be filling by each man’s Sol.16. (d)
 n
16 16  n ! 4 votes.
 Cn, r  = nC0 + nC1+ nC2+………+ nCn
 There is 5 ways for 1 place. r 0

18  n 17  n 16  n !  3 =5 x 5 x 5 x 5 = 625. ( we have 4 places) = 2n
16 16  n ! 4 Sol.9. (b) Sol.17. (b)
 18  n 17  n  12  n(n  3) 28
C2r  28C2r 4
Number of diagonals =
 306 17n 18n  n2  12 2 2r + 2r – 4 = 28
 n2  35n  294  0 Sol.10. (d) 4r = 32
BGBGBGB r=8

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Sol.18. (c) If captain is always included then we can choose at 2nd no. at 1st and 4th cannot repeat so no. of
As 'A' must be first letter of each word. 10 more players out of the remaining 14 players. possibilities = 8
Total number of words = 4! = 24 So at 3nd no. at 1st , 2nd and 4th cannot repeat so no. of
possibilities = 7
Sol.19. (a) P10  14!  1001
14

Number of words = 7C3 x4C2x 5! Total No’s = 8 x 8 x 7 x 5 = 2240

10!4!
7! 4! Sol.28. (b) Sol.36. (d)
 120    25200 Since sum of digits= 10 (which is not divisible
4!3! 2!2! First two words No. of words
AA 3!=6 by)
Sol.20. (d) AG 3!=6 No numbers can be formed
n n! AI 3!=6 Sol.37. (c)
Cr 1
r!n  r !
 AN 3!=6 Here, 15C3-nC3 = 445
1
n
C r 1 2  n!
 GA 3!=6 455 - nC3 = 445
2 GI 3!/2!=3 n
C3 = 10
(r  1)!n  r  1! GN 3!/2!=3 n=5
r 1 1 IA 3!=6
Sol.38. (a)
  IG 3!/2!=3
nr 2 IN 3!/2!=3
47
C3 + 47C4 + 48C3 + 49C3 + 50C3+ 51C3+ 52C3
 n  3r  2
48
……….(i) NA 3!=6 C4 + 48C3 + 49C3 + 50C3+ 51C3+ 52C3
49
n! Total = 54 C4 + 49C3 + 50C3+ 51C3+ 52C3
n
Cr 1 2 50
C4 + 50C3+ 51C3+ 52C3
 (r  1)!n  r  1! 2 It means 50th word will be starting with ‘NA’ 51
C4+ 51C3
n
Cr  2 3  n!
 th
N A A G I (49 Place)
3 52
C4
N A A I G (50th Place)
(r  2)!n  r  2! N A G A I (51st Place) Sol.39. (b)
r2 2 N A G I A (52nd Place) EQUATION -8 letters.
  Consonants – Q,T, N-3 letters.
n  r 1 3 N A I A G (53rd Place)
 2n  5r  8 ………(ii) N A I G A (54th Place) First letter of 8-letter word can be any of 3
by eq (i) and (ii) Sol.29. (c) consonants
n = 14 and r = 4 No. of digits to be filled at one’s place=3 Last letter of 8-letter word can be remaining 2
consonants.
Sol.21. (c) No. of digits to be filled at 10’s place=5 No.of
The middle 6-letters can be remaining 2
by above solution r = 4 digits to be filled at 100’s place=4
Consonants.
Sol.22. (b) No. of digits formed with zero at 100’s place = The middle 6-letters can be arranged in 6! ways.
n
Pr 1x2x4=8 So, number of different words = 3×2×6!
 r!  4! 24 Required no. of digits formed = 60 - 8 = 52
n
Cr = 6×720 = 4320
Sol.30. (d) Sol.40. (c)
Sol.23. (a) Each employee is eligible for 1 or more of the 10
C2 + 10C3 = 45 + 120 = 165
First letter can be put any 7 letters boxes=7 ways
Similarly, 2nd, 3rd, 4th and 5th letter be put in 7
tickets. Sol.41. (d)
No. of ways =20 x 20 x 20 = 8000. There will be 6 numbers
ways each respectively Sol.31. (a) Whose sum
7 x 7 x 7 x 7 x 7=75 No. of diagonals in a polygon 123+132+213+231+312+321=1332
Sol.24. (c) n  n  3 Sol.42. (a)
 44 
Number of ways that a cricket team of 11 2 C0+2[7C1+ 7C2+7C3+ 7C4+…... 7C6]+7C7
players can be made out of 15 players
 n2  3n  88  0 2.27 – 2 = 28- 2 = 256 – 2 = 254
15
C11 
15!
  n  11 n  8   0 Sol.43. (c)
11!4! Number of ways =
15 14 13 12 11! n = 11
8!8!
  1365 Sol.32. (b)  10!
11!1 2  3  4 4!6!
Let the given 4 digit decimal number is
10!
  8! 
2
Places after decimal can be filled in the
Sol.25. (c) following ways 4!6!
The given word is ‘NATION’ 10  9  8  7
7 8 9 9   8! 
2
Total number of words that can be formed form Total number of ways = 7 x 8 x 9 x 9 = 4536 4!
given word ‘NATION’ 10  9  8  7
Sol.33. (d)   8! 
2

6! 6  5  4  3  2! No. of ways in which 3 holidays travel tickets are 4  3  2 1

   360 to be given to 10 Sol.44. (a)
2! 2! Employees = 103 = 1000 PERMUTATION
Now numbers of word that can be formed from
given word NATION, so that all vowels never
Sol.34. (a) 11 letters and t is repeated 2 time.
Number of different messages that can be different permutations 11!
comes together.
represented by three 0’s and two 1’s is 
2!
 3! 
 360   4!   360   24  3 5! = 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3
 2!  10
3! x2! =19958400
 360  72  288 Option (a) is correct. Sol.45. (a)
Option (c) is correct
Sol.35. (c) To form a triangle, we need 3 points, 12 points
Sol.26. (d) are given So, 12P3 triangles can be formed.
1 n1 But, given that 7 points are on a straight line.
 n  r C n = Cn  Cn  1  n  1  n  2
n

r 0
Selecting 3 points from this set will not form a
from options 8 8 7 5 triangle.
n 2 at 4th place there are 5 possibilities 1,3,5,7,9 So, number of triangles formed 12C3–7C3
Cn1  n  2
at 1st 0 and digit and 4th place can not repeat so 12! 7!
Sol.27. (c) no. of possibilities = 8
 
3!9! 3!4!

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 12 1110 7  6  5 5 75 1 1 +
   220  35  185 
3  2 1 3  2 1 5 15 5.4! x  5(n  6)! 4x  4x  5! (n  6)
Sol.46. (a) 3 1
(a) A number divisible by 10 means the last digit 6.5.4! (n  6)!
is 0. 18 zeros in 75!
By solving above quadratic equation  n = 7
So, the remaining 3 digits can be arranged in 5 74
4×3×2 ways = 24 ways 5 14 Sol.70. (a)
Sol.47. (d) 2
Number between 100 and 1000 are 3-digit
numbers. It is given that the digits should not be 16 zeros in 74!
17 zeros not possible. There are 5 consonants and 2 vowel 2 vessel and
repeated.
2 consonants are selected as 5C22C2 = 10 types.
Number of given digits = 5 Sol.59. (c) These 4 letters can arrange as 4! =24
n
In a 3-digit number, first number can be arranged Cr × r! = nPr
Total possible cases = 10  24 = 240
in 5 ways. ⇒21× r! = 2520
Numbers that can be formed =5 × 4 × 3 = 60 ⇒r! = 120 ⇒ r = 5 Sol.71. (b)
n Number of straight lines = 20C2 = 20  19
Sol.48. (d) Pr = 2520
2
 190
(nCr + nCr-1) + (nCr-1 + nCr-2) n!
  2520 Number of triangles = 20C3 = 20  19  18 =1140
n+1
Cr + n+1Cr-1  n  5! 3  2 1
n+2
Cr ⇒n=7
Sol.49. (b) 8
C6 = 28 Sol.72. (d)
Total possible triangle = 10C3 = 120
Total numbers from between 2999 and 8001 are Sol.60. (d) Triangle with 1 common side of polygon = 60
no.of diagonals = n(n  3)  8  5  20
5001
Numbers with no repetition = 5x9x8x7 = 2520 Triangle with 2 common sides of polygon = 10
2 2
Numbers with at least 2 digits repeated = 5001- Triangle with no common side = 120 – 60 – 10 =
2520 = 2481 Sol.61. (c) 50
Sol.50. (a) C( 20, n + 2) = C(20 , n−2)
20
Cn−2 = 20Cn+2 Sol.73. (d)
3 digit number made from digit 3, 4 and 5 and 3n
C2n = 3nC2n–7
having all distinct digit=3!=6 and sum of such [nCr = nCn−r]
n − 2 + n + 2 = 20 2n + 2n–7 = 3n
numbers are 543 + 534 + 345 + 354 + 435 + 453
= 2664 n = 10 n=7
Sol.51. (c) Sol.62. (b) n
Cn–5 = 7C2 = 21
Number of 5 digits numbers with all distinct digit CVCV Sol.74. (c)
is same as filling of 5 vacant placed out of 10 Vowels can occupy 2 places and consonants may n
Cr = nCn–r
boxes. First digit of any number can be chosen in also occupy 2 places 51
C20 = 51C31, 51C21 = 51C30 and so on.
9 ways. Remaining 4 digits can be chosen in 2!2! = 4
Sum will be zero
remaining 9 digits in 9P4 ways. Sol.63. (c)
Total number of such number Each two circles will intersect at 2 places Sol.75. (a)
= 9 × 9 × 8 × 7 × 6 = 27216 so number of intersections = 2. 5C2 =20 3 1 5
7 9
Sol.52. (d) Sol.64. (d)
  
3 bowlers are selected among 5 bowlers in 5C3 sum of 1,2,3,4,5 is 15.
ways. so number formed by these integers is always 1  8  4 = 32
Remaining 8 player’s are selected from 12 divisible by 3. Sol.76. (d)
player’s in 12C8 ways. Sol.65. (c) Statement 1 is not correct for n=4
total number of ways = 12C8 ×5C3 one excluded so 7C5 = 21 Statement 2 is not correct for n=5
Sol.53. (d) Sol.66. (d) Sol.77. (b)
Given digits are 1,2,3,4 and 5 0(0!)+ 1(1!)+2(2!)+3(3!)+……..+10(10!) 5 player out of 7
Total number of 3-digit even numbers (1! - 0!)+ (2! - 1!)+ (3! - 2!)+…..+(11! - 10!) =7C5 = 21
= 4C2 × 2CI . 3  4 =11! - 1
  2  12 Sol.78. (b)
2 Sol.67. (c) O E O E O
Sol.54. (c) 3C2 2!  3!
Number of ways 4C16C4 + 4C26C3 + 4C36C2 +
4    3  2  6 = 36
C46C1
= (4) (15) + (6) (20) + (4) (15) + (1) (6) 3  6  5 = 90 Sol.79. (c)
= 60 + 120 + 60 + 6 = 246 Sol.68. (d) 30! + 35!
Sol.55. (b) 3, 5, 7, 9 digits are given to us, =30! (1+35  34  33  32  31)
A straight line can be formed by joining 2 points. We have to find 5 dight no’s from these digit 30! Contains
10
Total number of straight lines= C2 = 45 3579 7 times factor 5
Sol.56. (b)  5!  5  4  3  2! = 60 Sol.80. (b)
20
Cn+2 = 20Cn – 2 2! 2! 2(2  1) = 3!– 2!
n + 2 + n – 2 = 20 Here one position if left so one of the no. will 3(3  2  1) 4! –3!
n = 10 repeat itself
4(4  3  2 1) = 5!– 4!
Sol.57. (c) Sol.69. (a)
n
Cr + nCr+1 = n+1Cr+1 n
Ans so on
C4, nC5, nC6, are in AP (3!– 2!)+( 4! –3!)+( 5!– 4!)+…(10!– 9!)+2
= (47C3 + 47C4 )+ 48C3 + 49C3 + 50C3+ 51C3 2. nC5 = nC4+ nC6
= (48C4 + 48C3 ) + 49C3 + 50C3+ 51C3 = 10!
= (49C4 + 49C3 ) + 50C3+ 51C3 2. n!

n!

n! Sol.81. (a)
= (50C4 + 50C3) + 51C3 5! (n  5)! 4! (n  4)! 6! (n  6)! we have digits 1, 3, 5, 7, 9
= (51C3 + 51C3) = 52C4 number of cases = 5  5  5  5 = 625
Sol.58. (d) Sol.82. (c)

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for words starting from A , fix A at starting so So 4! + 5! + 6! + …… will give zero remainder 4th is man’s relative 3 men and women’s relative
number of words  10!  907200 because all are individually divisible by 8 3 womeni.e.4C0 3C33C34C0
4
So remainder will obtained only by 0! + 1! + 2! + C0 3C33C34C0 + 4C1 3C24C14C0 + 4C2 3C14C23C1 +
2!.2!. 4
3! = 10 C0 3C33C34C0 = 1 + 144 + 324 + 16 = 485
Sol.83. (d)
choose any 2 consonants and 2 vowels then
that is 2 Sol.100. (a)
arrange these 4 alphabets Sol.93. (d) there are total 12 points in the plane
5! = 120 , which is divisible by 24 so number of triangle formed by these points =
4
C2  4C2  4! = 864 12
So 5! + 6! + 7! + …… will give zero remainder C3 = 220
Sol.84. (c) if we take all three points of side PQ then
there are two type of words because all are individually divisible by 60
triangle cannot formed. (1)
CVCVCVCV or VCVCVCVC So remainder is 0! 1! + 2! + 3! + 4! = 34
if we take all three points of side QR then
( 4!  4! ) + (4!  4!) = 576 + 576 = 1152 Sol.94. (c) triangle cannot formed (4C3)
Sol.85. (b) 5! = 120 , which is divisible by 24 if we take all three points of side PR then triangle
all consonant together = 5! 4! = 2880 So 5! + 7! + …… will give zero remainder cannot formed (5C3)
because all are individually divisible by 24
Sol.86. (c) So remainder is 1! + 3! = 7
so number of triangles
if n is even number then nCr will be greatest if we = 220 – 1 - 4C3 - 5C3 = 205
take r = n/2 so statement 1 is correct. Sol.95. (c) Sol.101. (b)
if n is odd number then nCr will be greatest if we Number of intersection points by 10 circles
take r = (n+1)/2 or (n - 1)/2 so statement 2 is  210 C2  90 2 26
correct.
Sol.96. (b) 2 13
Sol.87. (c) even digits 0,2,4,6,8
m
C2 . nC2 = 60 2 6
at first place zero cannot come so
by hit and trial 2 3
number of ways = 4  5  5  5 = 500
if we take m = 2 then n cannot find
if we take m = 3 then n = 5 Sol.97. (d) 1
so m + n = 8 for divisibility of 4 , last 2 digits should be number of 2’s = 13 + 6 + 3 + 1 = 23
divisible by 4 i.e. = 223 = 4.221 = 4.87
Sol.88. (c) unit digit must be 2 so maximum value of k can be 7
12! and 12!
x y if we take last two digit 12 then there is 2 Sol.102. (c)
2!. 2!.2!.2!. possibilities 3512, 5312 max points of intersections of four straight lines
x = 4y if we take last two digit 32 then there is 2 = 4C2 = 6
Sol.89. (b) possibilities 1532, 5132 maximum intersection points of a circle and a
if we take last two digit 52 then there is 2 straight line = 2. 4C1 . 1C1 = 8
total possible numbers = 4  4  3  2 = 96
possibilities 1352, 3152 total points = 6 + 8 = 14
number greater than 5000 will start from 5 so
so there will be 6 cases.
number of cases = 1  4  3  2 = 24 Sol.103. (b)
24 is the 25 % of 96 Sol.98. (d) VCVCV
sum of number starting with 0 is 2220
Sol.90. (b) sum of number starting with 1 is 7998
2 consonants can arrange as 2!
25! + 1 is an odd number than cannot be divisible 3 vowels can arrange as 3! / 2!
sum of number starting with 4 is 25332 total required words = (3!  2!) / 2! = 6
by 26.
sum of number starting with 5 is 31110
6! + 1 = 721 that is divisible by 7
total = 64440
Sol.104. (c)
so statement 1 is incorrect and statement 2 is EQUATION
correct. Sol.99. (c) all vowels and consonants together
there are four possibilities
Sol.91. (d) 1st is man’s relative 3 women and women’s
VVVVVCCC + CCCVVVVV
number of 6 digit numbers ending with 2 or 4 = 4
relative 3 men i.e. 4C0 3C33C34C0 5!  3! + 3!  5! = 720 + 720 = 1440
 4  3  2  1  2 = 192
number of 6 digit numbers ending with 0= 5  4
2nd is man’s relative 1 man and 2 women , Sol.105. (d)
women’s relative 2 men and 1 woman a student choose (n – 2) courses out of n courses
 3  2  1  1 = 120 i.e.4C1 3C24C14C0 if 2 courses are compulsory
total = 192 + 120= 312 3rdis man’s relative 2 man and 1 women , = n – 2 Cn – 4 = (n – 2)(n – 3)/2
Sol.92. (c) women’s relative 1 men and 2 woman Sol.106. (a)
4! = 24 , which is divisible by 8 i.e.4C2 3C14C23C1 all four digits are odd and repetition is allowed =
5  5  5  5 = 625

SANDEEP SINGH BRAR Ph:-+91 9700900034 - 76 -