Question 1

A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is:

Accepted Answer

To ensure no bag is empty, exactly one bag must contain 2 books and the other two bags must contain 1 book each. The number of ways to choose 2 books out of 4 to be kept together is ^ 4 C_ 2 = 6 Now, we have 3 units (one pair of books and two single books) to be distributed into the 3 different bags. The number of ways to distribute these 3 units into 3 bags is 3! = 6 Total number of ways = 6 6 = 36 Answer: 36

Question 2

A building has ground floor and 10 more floors. Nine persons enter in a lift at the ground floor. The lift goes up to the 10^ th floor. The number of ways, in which any 4 persons exit at a floor and t

Accepted Answer

Total number of persons = 9 The lift goes up to the 10^ th floor but does not stop at the 1^ st and 2^ nd floors. Thus, the available floors for the persons to exit are from the 3^ rd to the 10^ th floor. Number of available floors = 8 The number of ways to select 4 persons out of 9 to form the first group is ^ 9 C_ 4 . The remaining 5 persons automatically form the second group. ^ 9 C_ 4 = 9 8 7 6 4 3 2 1 = 126 Now, these two distinct groups must exit at two different floors out of the 8 availa

Question 3

The number of 4-letter words, with or without meaning, each consisting of two vowels and two consonants that can be formed from the letters of the word INCONSEQUENTIAL, without repeating any letter, i

Accepted Answer

The given word is INCONSEQUENTIAL. The vowels present in the word are I, O, E, U, A. There are 5 distinct vowels. The consonants present in the word are N, C, S, Q, T, L. There are 6 distinct consonants. We need to form 4-letter words consisting of 2 vowels and 2 consonants without repeating any letter. This means all 4 letters in the formed word must be distinct. Number of ways to select 2 distinct vowels from 5 distinct vowels = ^ 5 C_ 2 = 5 4 2 = 10 Number of ways to select 2 distinct consona

Question 4

A box contains 5 blue, 6 yellow and 4 red balls. The number of ways, of drawing 8 balls containing at least two balls of each colour, is :

Accepted Answer

Let B, Y, and R be the number of blue, yellow, and red balls drawn respectively. We are given B + Y + R = 8 with B 2, Y 2, and R 2. The possible combinations for (B, Y, R) are: Case 1: (2, 2, 4) Case 2: (2, 3, 3) Case 3: (2, 4, 2) Case 4: (3, 2, 3) Case 5: (3, 3, 2) Case 6: (4, 2, 2) The number of ways for each case is calculated as follows: Case 1: ^ 5 C_ 2 ^ 6 C_ 2 ^ 4 C_ 4 = 10 15 1 = 150 Case 2: ^ 5 C_ 2 ^ 6 C_ 3 ^ 4 C_ 3 = 10 20 4 = 800 Case 3: ^ 5 C_ 2 ^ 6 C_ 4 ^ 4 C_ 2 = 10 15 6 = 900 Cas

Question 5

Two players A and B play a series of games of badminton. The player, who wins 5 games first, wins the series. Assuming that no game ends in a draw, the number of ways, in which player A wins the serie

Accepted Answer

Player A wins the series if A wins 5 games before B wins 5 games. The series can last for a minimum of 5 games and a maximum of 9 games. If player A wins the series in exactly n games, then A must win the n-th game, and A must win exactly 4 out of the first n-1 games. The number of ways this can happen is given by ^ n-1 C_ 4 . The possible values for n are 5, 6, 7, 8, and 9. Total number of ways for A to win the series is the sum of the number of ways A can win in 5, 6, 7, 8, or 9 games: Total w

Question 6

Let A=\ (a,b,c): a,b,c are non-negative integers and a+b+2c=22\ . Then n(A) is equal to:

Accepted Answer

The given equation is a + b + 2c = 22, where a, b, c are non-negative integers. This can be rewritten as a + b = 22 - 2c. Since a 0 and b 0, we must have 22 - 2c 0, which gives c 11. Since c is a non-negative integer, the possible values for c are 0, 1, 2, , 11. For a fixed value of c, the number of non-negative integer solutions to the equation a + b = 22 - 2c is given by (22 - 2c) + 1 = 23 - 2c. The total number of solutions n(A) is the sum of the number of solutions for each possible value of

Question 7

The number of ways of forming a queue of 4 boys and 3 girls such that all the girls are not together, is:

Accepted Answer

Total number of boys = 4 Total number of girls = 3 Total number of children = 7 Total number of ways to arrange 7 children in a queue = 7! = 5040 To find the number of ways where all 3 girls are together, consider the 3 girls as a single unit. Total number of units to arrange = 4 boys + 1 unit of girls = 5 units. Number of ways to arrange these 5 units = 5! = 120 Number of ways to arrange the 3 girls among themselves = 3! = 6 Total number of ways where all girls are together = 5! 3! = 120 6 = 72

Question 8

Let p_n denote the total number of triangles formed by joining the vertices of an n-side regular polygon. If p_ n+1 - p_n = 66, then the sum of all distinct prime divisors of n is:

Accepted Answer

The number of triangles formed by joining the vertices of an n-sided regular polygon is given by p_n = ^ n C_ 3 . Given that p_ n+1 - p_n = 66. ^ n+1 C_ 3 - ^ n C_ 3 = 66 Using the property ^ n+1 C_ r - ^ n C_ r = ^ n C_ r-1 , we get: ^ n C_ 2 = 66 n(n-1) 2 = 66 n(n-1) = 132 n^2 - n - 132 = 0 (n-12)(n+11) = 0 Since n must be a positive integer, n = 12. The prime factorization of 12 is 2^2 3. The distinct prime divisors of 12 are 2 and 3. The sum of all distinct prime divisors of n is 2 + 3 = 5.

Question 9

The number of seven-digit numbers, that can be formed by using the digits 1, 2, 3, 5 and 7 such that each digit is used at least once, is :

Accepted Answer

A seven-digit number is to be formed using the 5 digits 1, 2, 3, 5, and 7. Since each digit must be used at least once, the remaining 2 places can be filled by either repeating one digit two more times, or repeating two digits one more time each. This gives rise to two cases for the frequencies of the digits: Case 1: One digit appears 3 times and the other four digits appear 1 time each. The number of ways to choose the digit that appears 3 times is ^ 5 C_ 1 . The number of arrangements of these

Question 10

Three persons enter in a lift at the ground floor. The lift will go upto 10^ th floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not

Accepted Answer

Three persons must exit at three different floors. The lift goes to the 10th floor and does not stop at floors 1, 2, and 3. Available floors for exit: 4, 5, 6, 7, 8, 9, 10 (7 floors total). The number of ways to assign 3 persons to 3 different floors from 7 available floors is a permutation problem: P(7, 3) = 7 6 5 = 210. Alternatively, choose 3 floors from 7 and arrange 3 persons: ^ 7 P_ 3 = 35 6 = 210.

Question 11

Let S =\ 1,2,3,4,5,6,7,8,9\ . Let x be the number of 9 -digit numbers formed using the digits of the set S such that only one digit is repeated and it is repeated exactly twice. Let y be the number of

Accepted Answer

x counts 9-digit numbers using \ 1,2,3,4,5,6,7,8,9\ with exactly one digit repeated twice (8 distinct digits, one twice, one absent). Choose digit appearing twice: 9 ways. Choose absent digit: 8 ways. Arrange: 9! 2! ways. x = 9 8 9! 2! = 72 181440 = 13043520. y counts 9-digit numbers with exactly two digits each repeated twice (7 distinct digits, two appear twice, two absent). Choose 2 digits for repetition: 9 2 ways. Choose 2 absent digits: 7 2 ways. Arrange: 9! 2! 2! ways. y = 36 21 9! 4 = 756

Question 12

The largest value of n, for which 40^ n divides 60!, is

Accepted Answer

40^n = 2^ 3n 5^n. We need the highest power of 40 dividing 60!. Power of 5 in 60! = 60/5 + 60/25 = 12 + 2 = 14. Power of 2 in 60! = 60/2 + 60/4 + 60/8 + 60/16 + 60/32 = 30 + 15 + 7 + 3 + 1 = 56. We need n 14 and 3n 56 (i.e., n 18). The binding constraint is n 14, so the largest value is n = 14.

Question 13

The letters of the word "UDAYPUR" are written in all possible ways with or without meaning and these words are arranged as in a dictionary. The rank of the word "UDAYPUR" is

Accepted Answer

Letters of UDAYPUR: A, D, P, R, U, U, Y (U repeated twice). Words starting with a letter before U: letters A, D, P, R each give 6! 2! = 360 arrangements 4 360 = 1440. Fix U at position 1. Remaining: A, D, P, R, U, Y (all distinct). Words starting with UA: 5! = 120. Fix UD at positions 1-2. Remaining: A, P, R, U, Y. No letter before A, so fix UDA. Remaining: P, R, U, Y. Letters before Y: P, R, U give 3 3! = 18. Fix UDAY. Remaining: P, R, U. No letter before P, fix UDAYP. Remaining: R, U. Letter b

Question 14

The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0,1,2,5,9, if the repetition of the digits is allowed, is \_\_\_\_

Accepted Answer

(1) all different 5, 0, 1, 9 3 = 6 ways (2) 2 alike, 2 different 5, 0, 0, 1 3 ways 5, 1, 1, 2 3 ways 5, 2, 2, 0 3 ways 5, 2, 2, 9 3 ways 5, 5, 0, 2 6 ways 5, 5, 2, 9 6 ways 5, 1, 9, 9 3 ways (3) 3 alike, 1 different 5, 5, 5, 0 3 ways 5, 5, 5, 9 3 ways (4) 2 alike, 2 other alike 5, 5, 1, 1 3 ways Total ways = 42

Question 15

The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange, is

Accepted Answer

Distributing 16 identical oranges to 4 children with each getting at least 1. Let x_i 1 for i = 1,2,3,4 and x_1+x_2+x_3+x_4 = 16. Substituting y_i = x_i - 1 0: y_1+y_2+y_3+y_4 = 12. By stars and bars, number of ways = 12+3 3 = 15 3 = 15 14 13 6 = 455.

Question 16

Let S denote the set of 4-digit numbers a b c d such that a>b>c>d and P denote the set of 5 -digit numbers having product of its digits equal to 20. Then n( ~S )+n( P ) is equal to \_\_\_\_

Accepted Answer

n(S): Choose 4 distinct digits from \ 0,1, ,9\ and arrange in decreasing order. n(S) = 10 4 = 210. n(P): 5-digit numbers with digit product = 20 = 2^2 5. All digits 1 (else product = 0). Must include exactly one 5, remaining 4 digits multiply to 4. Case 1: \ 1,1,1,4,5\ : 5! 3! = 20 arrangements. Case 2: \ 1,1,2,2,5\ : 5! 2! 2! = 30 arrangements. n(P) = 50. Hence n(S) + n(P) = 210 + 50 = 260.

Question 17

The number of 4 -letter words, with or without meaning, which can be formed using the letters PQRPQRSTUVP, is \_\_\_\_.

Accepted Answer

Letters: P(3), Q(2), R(2), S(1), T(1), U(1), V(1) — 7 distinct letters. All distinct: 7 4 4! = 840. One pair + 2 distinct: choose repeating letter from \ P,Q,R\ (3 ways), choose 2 from remaining 6 ( 6 2 =15), arrange = 4!/2! = 12. Total = 3 15 12 = 540. Two pairs: choose 2 from \ P,Q,R\ ( 3 2 =3), arrange = 4!/(2!2!) = 6. Total = 18. One triple + 1 distinct: only P can repeat 3 times, choose 1 from 6, arrange = 4!/3! = 4. Total = 24. Answer = 840 + 540 + 18 + 24 = 1422.

Question 18

Let ABC be a triangle. Consider four points p _ 1 , p _ 2 , p _ 3 , p _ 4 on the side AB, five points p_ 5 , p_ 6 , p_ 7 , p_ 8 , p_ 9 on the side B C, and four points p_ 10 , p_ 11 , p_ 12 , p_ 13 on

Accepted Answer

Points: 4 on AB, 5 on BC, 4 on AC. For a valid pentagon, no 3 vertices can be collinear, so at most 2 points from each side. Only valid distribution (a, b, c) with a+b+c = 5 and each 2 is permutations of (1, 2, 2): (1,2,2): 4 1 5 2 4 2 = 4 10 6 = 240 (2,1,2): 4 2 5 1 4 2 = 6 5 6 = 180 (2,2,1): 4 2 5 2 4 1 = 6 10 4 = 240 Total = 240 + 180 + 240 = 660.

Question 19

The largest n N , for which 7^ n divides 101!, is :

Accepted Answer

Exponent of 7 in 101! is: 101 7 + 101 7^2 + 101 7^3 + = 14 + 2 = 16.

Question 20

The number of strictly increasing functions f from the set \ 1,2,3,4,5,6\ to the set \ 1,2,3, ., 9\ such that f(i) i for 1 i 6, is equal to :

Accepted Answer

For strictly increasing f, if we choose elements a_1 Since f(i) i always, condition f(i) i means f(i) i + 1 for all i. Let h(i) = f(i) - i. Then h(i) 1, h(6) 3, and h is non-decreasing. So h(i) \ 1, 2, 3\ for all i. Number of non-decreasing sequences of length 6 from \ 1, 2, 3\ = 8 2 = 28.

Permutation Combination — Previous Year Questions & Solutions