Practice UPSC CSAT previous year questions from 2011–2026 classified underPermutation and Combination. Review the original question, four options, correct answer and explanation for every PYQ.
28 questions • 2011–2026
How 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.
Treat each Q immediately followed by U as a block. With letters Q,U,E,U,E and the required adjacency, the number of distinct arrangements is 12 under the exam's intended counting of the two identical E's and two identical U's. Hence D.
Three numbers x, y, z are selected from the set of the first seven natural numbers such that x > 2y > 3z. How many such distinct triplets (x, y, z) are possible?
From x>2y>3z with values 1–7, z must be 1 and the feasible ordered triplets are four in number. Hence option D.
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?
In the worst case, all 34 non-red shoes can be drawn first. Then 10 red shoes could be drawn with each belonging to a different red pair; the 45th shoe must complete a red pair, so the maximum is 45.
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?
Solve a+4b+6c=25 in non-negative integers; counting the feasible combinations gives 19.
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. 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?
Exactly one misplaced letter is impossible; choosing two correct placements and swapping the other two gives 6 arrangements.
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?
There are 6 arrangements for the odd digits and 6 for the even digits: 36.
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?
The thousands digit is 1 and the remaining digits have 3!=6 arrangements; their sum is 7998.
In an examination, the maximum marks for each of the four papers P, Q, R and S are 100. Marks are integers. A student can score 99% in n different ways. What is n?
Total score is 396, so deficits from 100 sum to 4; the number of nonnegative solutions is C(7,3)=35.
A flag has to be designed with 4 horizontal strips using some or all of red, green and yellow. What is the number of ways so that no two adjacent stripes have the same colour?
First stripe has 3 choices and each subsequent stripe 2 choices: 3×2^3=24.
There are five persons P, Q, R, S and T, each 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?
If Task-2 is assigned to R, Task-1 can be S or T (2 choices), followed by 3! assignments of the remaining tasks: 12. The same applies when Task-2 is S, giving 24 total.
The digits 1 to 9 are arranged in three rows in such a way that each row contains three digits, and the number formed in the second row is twice the number formed in the first row, and the number formed in the third row is thrice the number formed in the first row. Repetition of digits is not allowed. If only three of the four digits 2, 3, 7 and 9 are allowed to use in the first row, how many such combinations are possible to be arranged in the three rows?
Checking the possible three-digit first-row numbers formed from three of 2,3,7,9 and requiring its double and triple to use the remaining digits exactly once leaves two valid arrangements.
In a tournament of Chess having 150 entrants, a player is eliminated whenever he loses a match. It is given that no match results in a tie/draw. How many matches are played in the entire tournament?
Each match produces exactly one elimination. To reduce 150 entrants to one winner requires 149 eliminations, hence 149 matches.
The 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?
A and E can occupy positions (1,4), (2,5), (4,1) or (5,2): four ordered placements. The other three letters can be arranged in 3! ways, giving 4 x 6 = 24.
A, 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?
There are 3 direct A-to-C routes and 3 x 4 = 12 routes through B. Total = 15.
There 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?
A valid PIN is a 3-element subset of 1 to 7 arranged in decreasing order, with every pair differing by at least 2. The valid subsets are 10 in total: (1,3,5), (1,3,6), (1,3,7), (1,4,6), (1,4,7), (1,5,7), (2,4,6), (2,4,7), (2,5,7), and (3,5,7). Therefore 10 attempts are sufficient in the worst case.
There are eight equidistant points on a circle. How many right-angled triangles can be drawn using these points as vertices and taking the diameter as one side of the triangle?
Every diameter is determined by choosing two opposite points; with eight equally spaced points there are 4 diameters. For each diameter, either of the remaining 6 points gives a right angle on the circle. Thus 4 x 6 = 24 triangles.
Consider the following statements in respect of a rectangular sheet of length 20 cm and breadth 8 cm: 1. It is possible to cut the sheet exactly into 4 square sheets. 2. It is possible to cut the sheet into 10 triangular sheets of equal area. Which of the above statements is/are correct?
Both statements are possible. The 20x8 rectangle can be cut into two 8x8 squares plus two 4x4 squares, exactly four square pieces. Its area is 160 sq cm; dividing the 20 cm length into five 4 cm strips and cutting each 4x8 rectangle along a diagonal gives ten triangles, each of area 16 sq cm.
One 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?
There are 5 choices for the initial vowel, 21 choices for the final consonant and 9 choices for the non-zero digit. Total = 5x21x9=945.
There 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?
Treat the 6 coffee and 3 tea cups by their types. There are C(9,3)=84 choices for the three tea positions. The only invalid arrangements are the 3 cases in which an entire row is tea, so 84-3=81.
What is the number of numbers of the form 0.XY, where X and Y are distinct non-zero digits?
X has 9 choices (1-9) and Y has 8 remaining non-zero choices. Thus there are 9x8=72 numbers.
Using 2, 2, 3, 3, 3 as digits, how many distinct numbers greater than 30000 can be formed?
A five-digit number greater than 30000 must begin with 3. The remaining four positions contain two 2s and two 3s, giving 4!/(2!2!)=6.
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?
The valid non-adjacent triples are {1,3,5}, {1,4,6}, {1,3,6}, and {2,4,6}: four combinations.
How many different sums can be formed with the denominations Rs. 50, Rs. 100, Rs. 200, Rs. 500 and Rs. 2,000 taking at least three denominations at a time?
Choose any 3, 4 or 5 of the five denominations: 10+5+1=16 possible combinations.
In a question paper there are five questions to be attempted and answer to each question has two choices — True (T) or False (F). It is given that no two candidates have given the answers to the five questions in an identical sequence. For this to happen, the maximum number of candidates is:
Each of five questions has two independent choices, giving 2^5=32 distinct answer sequences. Therefore at most 32 candidates can all have different sequences.
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
Choose the Principal in 2 ways. After that, choose 2 Vice-Principals from the remaining 5: C(5,2)=10. Total = 20, which is not listed.
A student has to opt for 2 subjects out of 5 subjects for a course, namely, Commerce, Economics, Statistics, Mathematics I and Mathematics II. Mathematics II can be offered only if Mathematics I is also opted. The number of different combinations of two subjects which can be opted is
There are C(4,2)=6 valid pairs excluding Mathematics II, plus the valid pair (Mathematics I, Mathematics II), giving 7.
There are 5 tasks and 5 persons. Task-1 cannot be assigned to either person-1 or person-2. Task-2 must be assigned to either person-3 or person-4. Every person is to be assigned one task. In how many ways can the assignment be done?
Task 2 has 2 choices. For each choice, Task 1 has 2 choices among the remaining eligible persons, and the remaining three tasks can be assigned in 3! ways: 2×2×6=24.
In a society it is customary for friends of the same sex to hug and for friends of opposite sex to shake hands when they meet. A group of friends met in a party and there were 24 handshakes. Which one among the following numbers indicates the possible number of hugs?
If there are m men and f women, handshakes give mf=24. The (4,6) split gives hugs C(4,2)+C(6,2)=6+15=21.
Attempt questions one by one with UPSC-style marking and immediate explanations.