UPSC Prelims PYQ › Quantitative Aptitude › Permutation, Combination & Probability
Permutation, Combination & Probability — UPSC CSAT Paper II Quantitative Aptitude previous year questions
20 free solved previous year questions on Permutation, Combination & Probability, papers from 2022 to 2026, drawn from UPSC CSAT, covering questions of the kind UPSC asks in CSAT Paper II. Every answer key has been checked again against published sources and solved independently, and each explanation is written fresh.
Read all 20 questions with answers
- #1UPSC CSAT 2026How many words can one form by shuffling the letters of the word QUEUE, if Q is always followed by U? The words thus formed need not necessarily have any meaning.
- A6
- B8
- C10
- D12
Answer: (D) 12Explanation
Q must sit right before a U. Treat 'QU' as one block.
The items to arrange are: block QU, U, E, E.
That is 4 items, with E repeated twice.
Number of arrangements = 4! ÷ 2! = 24 ÷ 2 = 12.
Each arrangement gives a different word.
Answer: 12, option (d).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #2UPSC CSAT 2026The top of a table is rectangular and its dimensions are 6′ × 10′. Two rectangular portions of the table top are painted in blue colour; both these portions have dimensions 2.5′ × 8′ and each of them has exactly two sides common with two edges of the table top. If the table is fixed to the ground and the remaining portion of the table top is painted in white, how many different patterns are possible when observed from above?
- A2
- B4
- C6
- D8
Answer: (B) 4Explanation
The table is 6′ wide and 10′ long. A 2.5′ × 8′ piece fits only with its 8′ side along the 10′ edge, since 8′ is more than 6′.
So each blue piece sits in one corner, touching two edges.
Two pieces of 8′ each placed at the two ends of the same 10′ edge would overlap, because 8 + 8 is more than 10. These cannot be two separate portions.
When on opposite long edges, 2.5 + 2.5 = 5 is less than 6, so they do not overlap.
Valid corner pairs: top-left with bottom-left, top-right with bottom-right, top-left with bottom-right, top-right with bottom-left.
The table is fixed, so these 4 patterns are different.
Answer: 4, option (b).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #3UPSC CSAT 2026There are four types of weights, namely 1 kg, 2 kg, 5 kg and 10 kg. What is the maximum number of different ways one can measure 20 kg, if at least eight but not more than eleven weights of 1 kg are to be used while measuring?
- A7
- B8
- C9
- D10
Answer: (B) 8Explanation
Let the number of 1 kg weights be a, where a is 8, 9, 10 or 11.
The rest, 20 – a, is made with 2 kg, 5 kg and 10 kg weights.
a = 8: 12 = 10+2, or 5+5+2, or six 2s. That is 3 ways.
a = 9: 11 = 5+2+2+2 only. That is 1 way.
a = 10: 10 = 10, or 5+5, or five 2s. That is 3 ways.
a = 11: 9 = 5+2+2 only. That is 1 way.
Total = 3 + 1 + 3 + 1 = 8, option (b).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #4UPSC CSAT 2023ABCD 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?
- A16
- B18
- C20
- D24
Answer: (C) 20Explanation
Six points give 6C3 = 20 triples.
A triple fails to form a triangle only if all three points are on one side.
No side carries more than two chosen points, so no triple is collinear.
Hence all 20 triples give triangles.
Answer: option (c).Source: UPSC CSAT 2023 question; answer key checked by independent computation
- #5UPSC CSAT 2023How 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?
- A12
- B18
- C36
- D72
Answer: (C) 36Explanation
Odd digits 1,1,3,3 fill the four odd positions in 4!/(2!2!) = 6 ways.
Even digits 2,2,4,4 fill the four even positions in 6 ways.
Total = 6 x 6 = 36 (confirmed by enumeration).
Answer: option (c).Source: UPSC CSAT 2023 question; answer key checked by independent computation
- #6UPSC CSAT 2023Raj 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?
- A27
- B36
- C44
- D45
Answer: (D) 45Explanation
A wearable pair needs one left and one right red shoe.
Worst case: he first draws all 17 x 2 = 34 white and black shoes.
Then he draws all 10 red shoes of the same foot (say all left): 10 more, total 44, still no pair.
The 45th shoe must be a red shoe of the other foot.
Answer: option (d).Source: UPSC CSAT 2023 question; answer key checked by independent computation
- #7UPSC CSAT 2023In 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?
- A18
- B19
- C20
- D21
Answer: (B) 19Explanation
Count non-negative solutions of a + 4b + 6c = 25, where a is the number of singles.
For each (b, c) with 4b + 6c <= 25, the singles are fixed.
Counting by number of sixes (0 to 4) gives 7, 5, 4, 2 and 1 cases.
Total = 19 (confirmed by enumeration).
Answer: option (b).Source: UPSC CSAT 2023 question; answer key checked by independent computation
- #8UPSC CSAT 2023There 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. It is possible that exactly one letter goes into an incorrect envelope.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?
- A1 only
- B2 only
- CBoth 1 and 2
- DNeither 1 nor 2
Answer: (B) 2 onlyExplanation
If three letters are in their right envelopes, the fourth envelope is forced to be right too, so exactly one wrong is impossible: statement 1 is false.
For exactly two correct, choose which two are right: 4C2 = 6, and the other two must be swapped (1 way).
So there are exactly 6 ways: statement 2 is true.
Answer: option (b).Source: UPSC CSAT 2023 question; answer key checked by independent computation
- #9UPSC CSAT 2023In 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?
- A16
- B17
- C23
- D35
Answer: (D) 35Explanation
99% of 400 is 396, so the four marks must sum to 396.
Let the shortfalls from 100 be non-negative integers summing to 4.
The number of ways to split a shortfall of 4 among 4 papers is 7C3 = 35.
A direct computer count of valid mark combinations also gives 35.
Answer: option (d).Source: UPSC CSAT 2023 question; answer key checked by independent computation
- #10UPSC CSAT 2023A 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?
- A12
- B18
- C24
- D36
Answer: (C) 24Explanation
First stripe: 3 colour choices.
Each later stripe must differ from the one above it, so it has 2 choices.
Total = 3 x 2 x 2 x 2 = 24.
Answer: option (c).Source: UPSC CSAT 2023 question; answer key checked by independent computation
- #11UPSC CSAT 2023There 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?
- A6
- B12
- C18
- D24
Answer: (D) 24Explanation
Task-2 goes to R or S: 2 ways.
Task-1 cannot go to P or Q or to the person who took Task-2, so it goes to the other of R/S or to T: 2 ways.
The remaining three persons take the three remaining tasks in 3! = 6 ways.
Total = 2 x 2 x 6 = 24.
Answer: option (d).Source: UPSC CSAT 2023 question; answer key checked by independent computation
- #12UPSC CSAT 2023What 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?
- A7998
- B8028
- C8878
- D9238
Answer: (A) 7998Explanation
Being below 2000 with no repeats, the thousands digit must be 1.
The other three digits 2, 3, 4 can be arranged in 3! = 6 ways: 1234, 1243, 1324, 1342, 1423, 1432.
Their sum is 7998.
Answer: option (a).Source: UPSC CSAT 2023 question; answer key checked by independent computation
- #13UPSC CSAT 2023What is the number of selections of 10 consecutive things out of 12 things in a circle taken in the clockwise direction?
- A3
- B11
- C12
- D66
Answer: (C) 12Explanation
A selection of 10 consecutive things in a circle is fixed by its starting point.
There are 12 possible starting points, and each gives a different set.
So there are 12 selections.
Answer: option (c).Source: UPSC CSAT 2023 question; answer key checked by independent computation
- #14UPSC CSAT 2022How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5?
- A8
- B12
- C16
- D24
Answer: (B) 12Explanation
Divisible by 5 with odd digits means the units digit is 5.
Hundreds digit: 4 choices from 1, 3, 7, 9. Tens digit: 3 remaining choices.
4 x 3 = 12.
Answer: option (b).Source: UPSC CSAT 2022 question; answer key checked by independent computation
- #15UPSC CSAT 2022The letters A, B, C, D and E are arranged in such a way that there are exactly two letters between A and E. How many such arrangements are possible?
- A12
- B18
- C24
- D36
Answer: (C) 24Explanation
A and E must occupy positions (1,4), (2,5) or the reverse, giving 2 position pairs x 2 orders = 4 ways.
The other three letters fill the remaining three places in 3! = 6 ways.
4 x 6 = 24.
Answer: option (c).Source: UPSC CSAT 2022 question; answer key checked by independent computation
- #16UPSC CSAT 2022A, B and C are three places such that there are three different roads from A to B, four different roads from B to C and three different roads from A to C. In how many different ways can one travel from A to C using these roads?
- A10
- B13
- C15
- D36
Answer: (C) 15Explanation
Via B: 3 x 4 = 12 ways.
Direct from A to C: 3 ways.
Total = 12 + 3 = 15.
Answer: option (c).Source: UPSC CSAT 2022 question; answer key checked by independent computation
- #17UPSC CSAT 2022There is a numeric lock which has a 3-digit PIN. The PIN contains digits 1 to 7. There is no repetition of digits. The digits in the PIN from left to right are in decreasing order. Any two digits in the PIN differ by at least 2. How many maximum attempts does one need to find out the PIN with certainty?
- A6
- B8
- C10
- D12
Answer: (C) 10Explanation
Choose three digits from 1-7 with every pair at least 2 apart; the order is then fixed (decreasing).
Listing: 531, 631, 641, 731, 741, 751, 642, 742, 752, 753, which is 10 (also C(5,3) = 10).
In the worst case 10 attempts are needed.
Answer: option (c).Source: UPSC CSAT 2022 question; answer key checked by independent computation
- #18UPSC CSAT 2022One non-zero-digit, one vowel and one consonant from English alphabet (in capital) are to be used in forming passwords, such that each password has to start with a vowel and end with a consonant. How many such passwords can be generated?
- A105
- B525
- C945
- D1050
Answer: (C) 945Explanation
The vowel is first and the consonant last, so the digit must be in the middle.
Choices: 5 vowels x 9 non-zero digits x 21 consonants.
5 x 9 x 21 = 945.
Answer: option (c).Source: UPSC CSAT 2022 question; answer key checked by independent computation
- #19UPSC CSAT 2022There are 9 cups placed on a table arranged in equal number of rows and columns out of which 6 cups contain coffee and 3 cups contain tea. In how many ways can they be arranged so that each row should contain at least one cup of coffee?
- A18
- B27
- C54
- D81
Answer: (D) 81Explanation
The cups form a 3 x 3 grid; choose 3 of 9 cells for tea: C(9,3) = 84 ways.
A row has no coffee only if all three tea cups fill that row: 3 ways.
84 - 3 = 81 (confirmed by enumeration).
Answer: option (d).Source: UPSC CSAT 2022 question; answer key checked by independent computation
- #20UPSC CSAT 2022What is the number of numbers of the form 0.XY, where X and Y are distinct non-zero digits?
- A72
- B81
- C90
- D100
Answer: (A) 72Explanation
X can be any of 9 non-zero digits.
Y can be any non-zero digit other than X, so 8 choices.
Total = 9 x 8 = 72.
Answer: option (a).Source: UPSC CSAT 2022 question; answer key checked by independent computation
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