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Permutation, Combination and Probability (UPSC) Part-I

📆 April 28, 2025

Total Questions: 51

41. A person has 4 coins each of different denomination. What is the number of different sums of money the person can form (using one or more coins at a time)? [2009]

Correct Answer: (b) 15
Solution:No. of different sums of money = any 1 coin at a time + any 2 coins + any 3 coins it all 4 coins
= ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄ = 4 + 6 + 4 + 1 = 15

42. How many numbers lie between 300 and 500 in which 4 comes only one time? [2009]

Correct Answer: (a) 99
Solution:

Case I : When 4 is at the hundredth place.
Remaining two places can be filled through any of the numbers 0 to 9 except 4 in 9 ways.
So, no. of ways = 1 × 9 × 9 = 81

Case II : When 4 is at the units or tens place and 3 is at the hundredth place. Here, 4 is at the units place, then tens place can be filled through any of the numbers 0 to 9 except 4 in 9 ways or else if 4 is at the tens place, then units place can be filled in 9 ways.
So, no. of ways = 1 × (9 + 9) = 1 × 18 = 18
Here, either case I or case II is possible. Hence, total no. of ways = 81 + 18 = 99

43. How many three-digit numbers can be generated from 1, 2, 3, 4, 5, 6, 7, 8, 9 such that the digits are in ascending order? [2009]

Correct Answer: (d) 84
Solution:Any 3 digits out of 9 can be selected in ⁹C₃ ways.
Now, these three digits can be arranged among themselves in ascending order in only 1 way.
Hence, total no. of ways = ⁹C₃ × 1 = 84

44. In a carrom board game competition, m boys and n girls (m > n > 1) of a school participate in which every student has to play exactly one game with every other student. Out of the total games played, it was found that in 221 games one player was a boy and the other player was a girl. Consider the following statements: [2009]

  1. The total number of students that participated in the competition is 30.
  2. The number of games in which both players were girls is 78.

Which of the statements given above is/are correct?

Correct Answer: (c) Both 1 and 2
Solution:Since in 221 games each boy plays exactly one game with each girl,
So, mn = 221 (1)
as, m > n > 1, so only m = 17 and n = 13 satisfies (1)
∴ Total no of students = m + n = 17 + 13 = 30
Number of games in which both players are girls = ¹³C₂ = 78

45. A question paper had ten questions. Each question could only be answered as True (T) or False (F). Each candidate answered all the questions. Yet, no two candidates wrote the answers in an identical sequence. How many different sequences of answers are possible? [2010]

Correct Answer: (d) 1024
Solution:Each question can be answered in 2 ways.
Hence, total no. of sequences = 2 × 2 × ........ 10 times
= 2¹⁰ = 1024

46. A person X has four notes of Rupee 1, 2, 5 and 10 denomination. The number of different sums of money she can form from them is [2010]

Correct Answer: (b) 15
Solution:Different sums of money can be formed by taking one, two, three and all the four notes together.
No. of different sums = ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄ = 4 + 6 + 4 + 1 = 15

47. When ten persons shake hands with one another, in how many ways is it possible? [2010]

Correct Answer: (d) 45
Solution:

First person can shake hand with the other 9 i.e., in 9 ways.
Second person can shake hand with the remaining 8 persons and so on.
∴ total no. of hands shaked = 9 + 8 + ........ + 2 + 1

=9(9+1)2 = 45 = 9(9+1)/2 = 45

48. In how many ways can four children be made to stand in a line such that two of them, A and B are always together? [2010]

Correct Answer: (b) 12
Solution:Take, A and B to be always together as a single entity.
Now, total no. of children = 4 - 2 + 1 = 3
These can be arranged in 3! ways and A, B can be arranged among themselves in 2! ways.
Hence, no. of arrangements such that A and B are always together = 3! × 2! = 3 × 2 × 2 = 12

49. There are 100 students in a particular class. 60% students play cricket, 30% students play football and 10% students play both the games. What is the number of students who play neither cricket nor football? [2011-II]

Correct Answer: (b) 20
Solution:

Given, n(U)=100n(U) = 100
Number of students who play cricket = 60 i.e. n(C)=60n(C) = 60
Number of students who play football = 30 i.e. n(F)=30n(F) = 30
Number of students who play both the games = 10
i.e. n(C∩F)=10n(C ∩ F) = 10
To find : n(C′∩F′)=?n(C' ∩ F') = ?

we know,

n(C∪F)=n(C)+n(F)−n(C∩F)=60+30−10=80n(C ∪ F) = n(C) + n(F) - n(C ∩ F) = 60 + 30 - 10 = 80 n(C′∩F′)=n(U)−n(C∪F)=100−80=20n(C' ∩ F') = n(U) - n(C ∪ F) = 100 - 80 = 20

50. A group of 630 children is seated in rows for a group photo session. Each row contains three less children than the row in front of it. Which one of the following number of rows is not possible? [2014-II]

Correct Answer: (d) 6
Solution:

Let no. of column = x, no. of rows = y

∴ xy = 630 − [3 × 1 + 3 × 2+........+3 × (y−1)]xy = 630

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