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CSAT · 43 questions

Counting, permutations & probability

Every UPSC CSAT question on this topic, 2016–2026, newest first. Tap an option to check yourself; the answer and explanation open below it.

Counting, permutations & probability questions per year: 2016: 4, 2017: 3, 2018: 4, 2019: 4, 2020: 2, 2021: 3, 2022: 8, 2023: 11, 2024: 1, 2025: 0, 2026: 3 Asked in 10 of 11 years · most in 2023 (11)

UPSC syllabus: “Basic numeracy (numbers and their relations, orders of magnitude, etc.) (Class X level), Data interpretation (charts, graphs, tables, data sufficiency etc. — Class X level);” See the full syllabus →

CSAT 2023 · Q4

Medium

Raj has ten pairs of red, nine pairs of white and eight pairs of black shoes in a box. If he randomly picks shoes one by one (without replacement) from the box to get a red pair of shoes to wear, what is the maximum number of attempts he has to make?

Answer & explanation

Answer: (d) 45

In the worst case Raj first takes out every white and black shoe, and then every red shoe for one foot. A pair he can wear needs one left and one right shoe, so the pair is certain only on the next draw: 18 + 16 + 10 + 1 = 45.

  1. The box has 20 red shoes (10 left, 10 right), 18 white shoes and 16 black shoes.
  2. Worst case: all 18 white and all 16 black shoes come out first: 34 draws and still no red pair.
  3. Next, all 10 red shoes of one foot (say all left shoes) come out: 44 draws and still no pair to wear.
  4. The 45th draw has to be a red right shoe, which completes a red pair. Maximum attempts = 34 + 10 + 1 = 45.
  5. Check: ignoring left and right would give 34 + 2 = 36, the trap option; a wearable pair needs one shoe of each foot.

Remember · Worst-case questions: use up every unhelpful draw first, including same-foot or same-type items, then add one.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q5

Medium

In how many ways can a batsman score exactly 25 runs by scoring single runs, fours and sixes only, irrespective of the sequence of scoring shots?

Answer & explanation

Answer: (b) 19

Count the combinations of sixes, fours and singles that add up to 25; the order of shots does not matter. Fix the number of sixes, count how many fours can fit, and let singles make up the rest: 7 + 5 + 4 + 2 + 1 = 19.

  1. Let the batsman hit x sixes, y fours and z singles: 6x + 4y + z = 25. Once x and y are chosen, z is fixed.
  2. x = 0: 4y ≤ 25, so y = 0 to 6 → 7 ways.
  3. x = 1: 4y ≤ 19, so y = 0 to 4 → 5 ways.
  4. x = 2: 4y ≤ 13, so y = 0 to 3 → 4 ways.
  5. x = 3: 4y ≤ 7, so y = 0 or 1 → 2 ways.
  6. x = 4: 4y ≤ 1, so y = 0 → 1 way. Total = 7 + 5 + 4 + 2 + 1 = 19.

Remember · For 'irrespective of order' counts, fix the biggest unit, count options for the next; the smallest unit fills the rest automatically.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q6

Medium

There are four letters and four envelopes and exactly one letter is to be put in exactly one envelope with the correct address. If the letters are randomly inserted into the envelopes, then consider the following statements:

  1. 1.It is possible that exactly one letter goes into an incorrect envelope.
  2. 2.There are only six ways in which only two letters can go into the correct envelopes.

Which of the statements given above is/are correct?

Answer & explanation

Answer: (b) 2 only

If three letters are in their correct envelopes, the only envelope left belongs to the fourth letter, so exactly one wrong letter is impossible. For exactly two correct, choose the two correct letters in 6 ways and swap the other two in 1 way: 6 ways.

  1. Statement 1: if 3 letters are correct, the 4th letter is left with only its own envelope, so it is correct too. Exactly one wrong letter cannot happen.
  2. Statement 2: choose the 2 letters that go correctly: C(4, 2) = 6 ways.
  3. The other 2 letters must both be wrong, which is possible in only 1 way (they swap envelopes).
  4. Ways = 6 × 1 = 6, so statement 2 is correct.
  • ✗ 1. With three letters correctly placed, the last envelope is the fourth letter's own; one misplaced letter alone is impossible.
  • ✓ 2. C(4, 2) = 6 choices of the two correct letters, and the remaining two can be wrong in just one way (a swap): 6 ways.

Remember · Exactly one item misplaced is never possible. 'Exactly k correct' = C(n, k) × ways to put all the rest wrong.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q19

Medium

How many distinct 8-digit numbers can be formed by rearranging the digits of the number 11223344 such that odd digits occupy odd positions and even digits occupy even positions?

Answer & explanation

Answer: (c) 36

The four odd positions must hold 1, 1, 3, 3 and the four even positions must hold 2, 2, 4, 4. Each set can be arranged in 4!/(2! × 2!) = 6 ways, so there are 6 × 6 = 36 numbers.

  1. Odd digits 1, 1, 3, 3 fill the 4 odd positions; even digits 2, 2, 4, 4 fill the 4 even positions.
  2. Arrangements of 1, 1, 3, 3 = 4!/(2! × 2!) = 24/4 = 6.
  3. Arrangements of 2, 2, 4, 4 = 6 in the same way.
  4. Total = 6 × 6 = 36.

Remember · Arrangements with repeated items = n!/(p! × q! …); choices for independent slots multiply.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q30

Easy

ABCD is a square. One point on each of AB and CD; and two distinct points on each of BC and DA are chosen. How many distinct triangles can be drawn using any three points as vertices out of these six points?

Answer & explanation

Answer: (c) 20

Three points fail to make a triangle only if they lie on one straight line. No side of the square carries more than two of the six points, so no three are in a line and every choice of three points gives a triangle: C(6, 3) = 20.

  1. Points: 1 on AB, 1 on CD, 2 on BC and 2 on DA — six in all.
  2. A set of three points forms no triangle only when all three lie on one side; no side has more than two points.
  3. Number of triangles = C(6, 3) = (6 × 5 × 4)/(3 × 2 × 1) = 20.

Remember · Triangles from points = C(n, 3) minus collinear triples; check each line for three or more points.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q36

Medium

A box contains 14 black balls, 20 blue balls, 26 green balls, 28 yellow balls, 38 red balls and 54 white balls. Consider the following statements:

  1. 1.The smallest number n such that any n balls drawn from the box randomly must contain one full group of at least one colour is 175.
  2. 2.The smallest number m such that any m balls drawn from the box randomly must contain at least one ball of each colour is 167.

Which of the above statements is/are correct?

Answer & explanation

Answer: (c) Both 1 and 2

For a full group, the unluckiest draw takes every colour one ball short of complete: 13 + 19 + 25 + 27 + 37 + 53 = 174, so 175 balls guarantee a full group. For one ball of each colour, the unluckiest draw takes every ball except the smallest colour: 180 − 14 = 166, so 167 balls guarantee all six colours.

  1. Total balls = 14 + 20 + 26 + 28 + 38 + 54 = 180.
  2. Statement 1: the worst case leaves every colour one short: 13 + 19 + 25 + 27 + 37 + 53 = 174 balls with no full group. The next ball completes one, so n = 175. Correct.
  3. Statement 2: the worst case is all balls except the smallest colour (14 black): 180 − 14 = 166 balls without black. The next ball must be black, so m = 167. Correct.
  • ✓ 1. 174 balls can still leave every colour one ball short of full; the 175th ball must complete a group.
  • ✓ 2. 166 balls can all be non-black; the 167th must be black, giving all six colours.

Remember · Guarantee questions: build the unluckiest draw that still fails, then add one.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q47

Medium

What is the sum of all 4-digit numbers less than 2000 formed by the digits 1, 2, 3 and 4, where none of the digits is repeated?

Answer & explanation

Answer: (a) 7998

Numbers below 2000 must start with 1, and the other three places take 2, 3 and 4 in 3! = 6 orders. The thousands digits give 6 × 1000 = 6000; in each other place each of 2, 3 and 4 appears twice, adding 18 × 111 = 1998. The total is 7998.

  1. Below 2000 with the digits 1, 2, 3, 4 used once each, the first digit must be 1; 2, 3, 4 fill the rest in 3! = 6 ways.
  2. Thousands place: 6 × 1000 = 6000.
  3. In each of the hundreds, tens and units places, each of 2, 3, 4 appears 2 times: (2 + 3 + 4) × 2 = 18.
  4. Their contribution = 18 × (100 + 10 + 1) = 18 × 111 = 1998.
  5. Total = 6000 + 1998 = 7998.

Remember · Sum of numbers formed by permutations: count how often each digit sits in each place, then multiply by the place values (111…).

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q48

Easy

What is the number of selections of 10 consecutive things out of 12 things in a circle taken in the clockwise direction?

Answer & explanation

Answer: (c) 12

Going clockwise, a block of 10 consecutive things is fixed by its starting point, and each of the 12 things can be the start. So there are 12 selections.

  1. Taken clockwise, a run of 10 consecutive things is decided by where it starts.
  2. Each of the 12 things can be the starting point, and different starts give different sets.
  3. Number of selections = 12.
  4. Check: each selection leaves out 2 neighbouring things, and a circle of 12 has exactly 12 neighbouring pairs.

Remember · In a circle of n items, the number of blocks of k consecutive items (k < n) is n.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q65

Medium

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?

Answer & explanation

Answer: (d) 35

99% of 400 is 396, so the student loses exactly 4 marks in all across the four papers. Sharing 4 lost marks among 4 papers (0 or more each) can be done in C(7, 3) = 35 ways.

  1. Total maximum = 4 × 100 = 400; 99% of 400 = 396.
  2. Let the marks lost in P, Q, R, S be w, x, y, z (each 0 or more) with w + x + y + z = 4; no paper can exceed 100.
  3. Number of solutions = C(4 + 3, 3) = C(7, 3) = 35.
  4. Check by cases: (4,0,0,0) type 4 ways, (3,1,0,0) 12, (2,2,0,0) 6, (2,1,1,0) 12, (1,1,1,1) 1 → 35.

Remember · Count marks lost instead of marks scored; n identical units in r boxes can be shared in C(n + r − 1, r − 1) ways.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q66

Easy

A flag has to be designed with 4 horizontal stripes using some or all of the colours red, green and yellow. What is the number of different ways in which this can be done so that no two adjacent stripes have the same colour?

Answer & explanation

Answer: (c) 24

The top stripe can take any of 3 colours, and each later stripe any colour except the one just above it (2 choices). So there are 3 × 2 × 2 × 2 = 24 flags.

  1. Stripe 1: 3 choices.
  2. Stripes 2, 3 and 4: each must differ from the stripe above it, so 2 choices each.
  3. Total = 3 × 2 × 2 × 2 = 24.
  4. 'Some or all of the colours' allows two-colour flags such as red-green-red-green; they are already included.

Remember · For 'no two adjacent alike', choose the first freely, then (colours − 1) for each next item.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q68

Medium

There are five persons P, Q, R, S and T each one of whom has to be assigned one task. Neither P nor Q can be assigned Task-1. Task-2 must be assigned to either R or S. In how many ways can the assignment be done?

Answer & explanation

Answer: (d) 24

Task-2 goes to R or S (2 ways). Task-1 cannot go to P or Q or to whoever took Task-2, so it goes to the other of R and S or to T (2 ways). The remaining three tasks go to the remaining three persons in 3! = 6 ways: 2 × 2 × 6 = 24.

  1. Five persons, five tasks, one task each.
  2. Task-2: R or S → 2 ways.
  3. Task-1: not P, not Q, and not the person given Task-2 → 2 ways (the other of R and S, or T).
  4. Tasks 3, 4 and 5 go to the 3 persons left: 3! = 6 ways.
  5. Total = 2 × 2 × 6 = 24.

Remember · Fill the most restricted positions first, then arrange the rest freely.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

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