Permutation and Combination for CSAT: PYQs Explained

Permutation and combination is the mathematics of counting arrangements without listing them one by one. A permutation counts arrangements where the order matters, like who sits in which chair. A combination counts selections where the order does not matter, like which three books you pick from a shelf. In CSAT Paper II, a qualifying paper (roughly 33 percent clears it; marks do not add to your merit rank), this topic appears almost every year, and the entire subject rests on two rules and two formulae.

The two counting rules everything rests on

The multiplication rule says: if a task happens in stages, and stage 1 has m options and stage 2 has n options independently, the total number of ways is m times n. The addition rule says: if you can do a task either one way or another (mutually exclusive alternatives), the total is the sum. The roads question is the classic combined example.

Worked example (UPSC CSE 2022). 3 roads run from A to B, 4 roads from B to C, and 3 roads directly from A to C. Ways to travel from A to C? Two mutually exclusive alternatives: go direct (3 ways) or go via B (3 times 4 = 12 ways, by the multiplication rule). Addition rule: 3 + 12 = 15 ways, option (c).

Worked example (UPSC CSE 2015). There are 5 tasks and 5 persons; task 1 cannot go to person 1 or 2, and task 2 must go to person 3 or 4; every person gets exactly one task. Fix task 2 first (2 choices: person 3 or 4). If task 2 goes to person 3, task 1 has 2 choices left (persons 4, 5) and the remaining 3 persons fill the remaining 3 tasks in 3 times 2 times 1 = 6 ways: 2 times 1 times 6 = 12. The same count, 12, holds when task 2 goes to person 4. Total = 12 + 12 = 24, option (c). The lesson: handle the most restricted stage first, then multiply.

Permutations: order matters

A permutation is an ordered arrangement of r distinct objects chosen from n. The count is nPr = n! / (n minus r)!, where n! (n factorial) means n times (n minus 1) times ... times 1. The multiplication rule gives the intuition: n choices for the first slot, n minus 1 for the second, down to n minus r + 1 for the r-th, and that product is exactly nPr.

Worked example (UPSC CSE 2018). Six different colours are available; three wooden blocks of a winners' stand must be painted so that no two blocks share a colour. Ways? First block: 6 colours; second: 5; third: 4. Multiply: 6 times 5 times 4 = 120, option (a). Order matters here because the three blocks are different objects, even though the question never says the word permutation.

Worked example (UPSC CSE 2015). A student must opt for 2 subjects from Commerce, Economics, Statistics, Mathematics I and Mathematics II, but Mathematics II can be taken only with Mathematics I. All pairs: C(5, 2) = 10. Forbidden pairs: Mathematics II with each of Commerce, Economics, Statistics = 3. Valid = 10 minus 3 = 7, option (c). Count the total, subtract the forbidden: a standard UPSC move.

Combinations: order does not matter

A combination is a selection of r objects from n where the order is irrelevant. The count is nCr = n! / (r! times (n minus r)!). The division by r! corrects the overcounting: each group of r objects was counted r! times as a permutation, but a selection should be counted once. Two identities save time: nCr = nC(n minus r) (choose the r you take, or the n minus r you leave), and nC1 = n.

Worked example (UPSC CSE 2022). 9 cups in a 3-by-3 grid; 6 contain coffee and 3 contain tea. In how many arrangements does each row contain at least one cup of coffee? Place the 3 tea cups: C(9, 3) = 84 total arrangements. A row fails only if all 3 tea cups sit in it: 3 choices of row. Valid = 84 minus 3 = 81, option (d). Again: total minus forbidden.

Worked example (UPSC CSE 2015). 2 candidates are eligible for 1 Principal post; all 6 are eligible for 2 Vice-Principal posts. Combinations of selectees? Principal: C(2, 1) = 2. Vice-Principals from the remaining 5: C(5, 2) = 10. Total = 2 times 10 = 20. Since 20 is not among the options (4, 12, 18), the answer is none of the above, option (d). UPSC does plant such options; trust your arithmetic.

The non-consecutive selection trick

UPSC loves the restriction no two chosen together. Choosing r objects from n in a row with none adjacent equals C(n minus r + 1, r). The trick behind it: shrink each chosen object together with the forced gap after it, turning the problem into choosing r slots from n minus r + 1.

Worked example (UPSC CSE 2021). 6 persons stand in a row; another person must shake hands with 3 of them but never with two consecutive persons. Ways? C(6 minus 3 + 1, 3) = C(4, 3) = 4, option (b). Verify by hand: the valid triples are {1,3,5}, {1,3,6}, {1,4,6}, {2,4,6}, exactly four.

Reverse reasoning: the total tells you the split

Some questions give you a total and ask which option could produce it. List the factor pairs of the total, compute the answer for each pair, and keep the one that appears among the options.

Worked example (UPSC CSE 2015). Same-sex friends hug, opposite-sex friends shake hands; a party had 24 handshakes. Which could be the number of hugs? If there are m men and w women, handshakes = m times w = 24. Factor pairs: (1, 24), (2, 12), (3, 8), (4, 6). Hugs = C(m, 2) + C(w, 2): pair (4, 6) gives 6 + 15 = 21, and 21 is the only option present, so the answer is option (c).

Formula cheat sheet

Traps UPSC sets

• Permutation versus combination misread. Painting three different blocks is a permutation (6 times 5 times 4); forming handshake pairs is a combination. Ask: does swapping the chosen objects create something different?
• Multiply when you should add. Direct road versus via-B are alternatives: add (3 + 12), not multiply. Use multiply only for stages that happen in sequence.
• Forgetting the forbidden arrangements. The subjects and cups questions both fall to total minus forbidden; listing valid cases directly is slow and error-prone.
• Choosing from those left, not from all. In the Principal question the Vice-Principals are chosen from the remaining 5, not 6; the Principal's selection consumes one candidate.
• Order of constraints. Fix the most restricted choice first (task 2 with only 2 options); starting with the free choices multiplies your casework.

Speed tactics

• Shrink with nCr = nC(n minus r). C(9, 7) is C(9, 2) = 36; always work with the smaller of r and n minus r.
• Expand factorials partially. 9 times 8 times 7 over 3 times 2 times 1: cancel before you multiply, never compute 9! fully.
• Total minus forbidden first. For at least one and with-restriction questions, the complement is almost always the shorter computation.
• Eliminate by divisibility. If the stem's structure forces a multiple of 3 and only one option qualifies, you are done without full arithmetic.

Key Terms

• Permutation: an ordered arrangement of objects; swapping two objects gives a different permutation.
• Combination: a selection of objects where order is irrelevant; each selection is counted once.
• Factorial (n!): the product of all positive integers up to n; counts the arrangements of n distinct objects.
• Arrangement: a placement of objects into distinct positions or slots, where which object goes where matters.
• Selection: a group of objects chosen together, where the order of choosing carries no meaning.
• Multiplication principle: if a task has sequential independent stages, total ways = product of the options at each stage.
• Addition principle: if a task can be done in mutually exclusive alternative ways, total ways = sum of the counts.
• Overcounting: counting the same arrangement multiple times; corrected by dividing, e.g. nCr divides nPr by r!.
• Identical objects: objects that cannot be told apart; swapping them creates no new arrangement, so divide the count accordingly.
• Forbidden (excluded) arrangements: the rule-breaking cases subtracted from the total in a total-minus-forbidden computation.

Practice questions