Permutation, Combination & Probability MCQs for UPSC Prelims

94 practice questions on Permutation, Combination & Probability from the Quantitative Aptitude section of the UPSC Prelims syllabus. 94 come with a written explanation and 52 are actual previous year questions. Try the sample set below - the answer stays hidden until you ask for it.

19 Easy 51 Medium 24 Hard 52 from past papers

Sample questions

Q1
Previous year question hard

While writing all the numbers from 700 to 1000, how many numbers occur in which the digit at hundred's place is greater than the digit at ten's place, and the digit at ten's place is greater than the digit at unit's place?

  1. A 61
  2. B 64
  3. C 85
  4. D 91
Show answer and explanation

Correct answer: C - 85

We need three-digit numbers with strictly decreasing digits (hundreds > tens > units). The number 1000 has digits 1,0,0,0 and does not qualify, so the hundreds digit can only be 7, 8 or 9. For a fixed hundreds digit H, the tens and units digits must be two distinct digits both less than H, arranged in decreasing order; choosing any 2 distinct digits from {0,1,...,H-1} fixes their order, giving C(H, 2) numbers. For H = 7: C(7,2) = 21. For H = 8: C(8,2) = 28. For H = 9: C(9,2) = 36. Total = 21 + 28 + 36 = 85. Hence (c) 85.

Q2
Previous year question hard

Each face of a cube can be painted in black or white colours. In how many different ways can the cube be painted?

  1. A 9
  2. B 10
  3. C 11
  4. D 12
Show answer and explanation

Correct answer: B - 10

Count distinct colourings of a cube with 2 colours where two colourings are the same if one can be rotated into the other. Group by number of black faces: 0 black (1 way), 6 black (1 way); 1 black (1 way, all single faces are equivalent by rotation), 5 black (1 way); 2 black (2 ways, the black faces are either adjacent or opposite), 4 black (2 ways, by symmetry with 2 black); 3 black (2 ways, the three black faces either share a common vertex or form a band/strip). Adding: 1 + 1 + 1 + 1 + 2 + 2 + 2 = 10. (This matches Burnside's count (1/24)(2^6 + 6*2^3 + 3*2^4 + 8*2^2 + 6*2^3) = 240/24 = 10.) So the answer is (b) 10.

Q3
Previous year question easy

Study the following figure. A person goes from A to B always moving to the right or downwards along the lines. The figure is a square grid of 2 x 2 cells, with A at the top-left corner and B at the bottom-right corner. How many different routes can he adopt? Select the correct answer from the codes given below:

  1. A 4
  2. B 5
  3. C 6
  4. D 7
Show answer and explanation

Correct answer: C - 6

To go from A (top-left) to B (bottom-right) of a 2 x 2 grid of cells, the person must make exactly 2 moves to the right (R) and 2 moves downward (D), in any order. The number of distinct sequences of these 4 moves is the number of ways to choose which 2 of the 4 steps are 'right', that is 4C2 = 4! / (2! x 2!) = 6. So there are 6 different routes.

Q4
Previous year question hard

In an examination, the maximum marks for each of the four papers namely P, Q, R and S are 100. Marks scored by the students are in integers. A student can score 99% in n different ways. What is the value of n?

  1. A 16
  2. B 17
  3. C 23
  4. D 35
Show answer and explanation

Correct answer: D - 35

99% of the 400 total is 396, so the four integer scores must add to 396, i.e. the shortfalls from 100 in the four papers must add to 4 with each shortfall between 0 and 100. The number of non-negative integer solutions of d1 + d2 + d3 + d4 = 4 is C(4 + 3, 3) = C(7, 3) = 35 (the cap of 100 never binds since each shortfall is at most 4). Enumerating score patterns confirms it: (100,100,100,96) in 4 ways, (100,100,99,97) in 12, (100,100,98,98) in 6, (100,99,99,98) in 12, (99,99,99,99) in 1, totalling 35. So (d) is correct; 16, 17 and 23 come from missing some of these distributions.

Q5
Previous year question medium

How many five-digit prime numbers can be obtained by using all the digits 1, 2, 3, 4 and 5 without repetition of digits?

  1. A Zero
  2. B One
  3. C Nine
  4. D Ten
Show answer and explanation

Correct answer: A - Zero

Any five-digit number formed from the digits 1, 2, 3, 4, 5 uses each digit exactly once, so its digit sum is always 1 + 2 + 3 + 4 + 5 = 15. Since 15 is divisible by 3, every such arrangement is divisible by 3 (and is greater than 3), so none can be prime. Therefore the count of primes is Zero.

Q6
Previous year question medium

A selection is to be made for one post of Principal and two posts of Vice-Principal. Amongst the six candidates called for the interview, only two are eligible for the post of Principal while they all are eligible for the post of Vice-Principal. The number of possible combinations of selectees is

  1. A 4
  2. B 12
  3. C 18
  4. D None of the above
Show answer and explanation

Correct answer: D - None of the above

Choose the Principal first. Only 2 of the 6 candidates are eligible for Principal, so there are 2 ways. Once the Principal is fixed, that person is no longer available, leaving 5 candidates for the two identical Vice-Principal posts. The two Vice-Principals are an unordered pair, so the count is 5C2 = 10. Total combinations = 2 x 10 = 20. Since 20 does not appear among options (a) 4, (b) 12, or (c) 18, the correct choice is None of the above. The official UPSC 2015 key marks this answer as (d).

Q7
Previous year question easy

How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5?

  1. A 8
  2. B 12
  3. C 16
  4. D 24
Show answer and explanation

Correct answer: B - 12

A number divisible by 5 must end in 0 or 5; since every digit must be odd, the units digit is 5. The hundreds and tens digits must come from the remaining odd digits {1, 3, 7, 9} without repetition: 4 choices for the hundreds place and 3 for the tens place, giving 4 x 3 = 12 numbers, option (b). 8 undercounts the choices, while 16 and 24 ignore the no-repetition or odd-digit constraints.

Q8
Previous year question medium

There are 6 persons arranged in a row. Another person has to shake hands with 3 of them so that he should not shake hands with two consecutive persons. In how many distinct possible combinations can the handshakes take place?

  1. A 3
  2. B 4
  3. C 5
  4. D 6
Show answer and explanation

Correct answer: B - 4

Number the six persons 1 to 6 in the row. We need 3-element subsets with no two consecutive numbers. The count of ways to pick k non-consecutive items from n in a row is C(n-k+1, k) = C(4, 3) = 4. Listing them confirms it: {1,3,5}, {1,3,6}, {1,4,6}, {2,4,6}. So the answer is 4, option (b). Option (a) 3 misses one valid set, while option (c) 5 and option (d) 6 include sets such as {1,2,4} or {2,4,5} that contain adjacent persons.

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