Permutation or Combination? The 5-Second Test for JEE Counting Problems

Ask ten students whether "choose 3 people from 5" is a permutation or a combination problem, and half will guess. Guessing works until the question is worded differently, and then marks go. There is a clean test that removes the guesswork, and it takes five seconds.

Quick Summary

• A combination counts selections where order does not matter: nCr = n! / (r! × (n − r)!).

• A permutation counts arrangements where order matters: nPr = n! / (n − r)!.

• The two are linked by nPr = nCr × r!, because every selection can be arranged in r! ways.

• The deciding test: swap two chosen items. If the outcome is now a different outcome, use permutations; if it is the same, use combinations.

• This matters for JEE Main and JEE Advanced (Mathematics). NEET does not test it.

Start from what you already know

Suppose you have three friends, A, B and C, and two tickets. If the tickets are identical, giving them to A and B is the same as giving them to B and A. That is 3 outcomes: AB, AC, BC. If instead the tickets are for seat 1 and seat 2, then A in seat 1 and B in seat 2 differs from B in seat 1 and A in seat 2. That is 6 outcomes: AB, BA, AC, CA, BC, CB.

Notice that 6 = 3 × 2. Every unordered pair produced exactly 2 ordered pairs, and 2 is 2!. This is the entire relationship between the two ideas, and it is why the formulas look so alike.

Why the formulas are what they are

To arrange r items out of n, fill positions one by one: n choices for the first, (n − 1) for the second, and so on down to (n − r + 1) for the r-th. Multiplying gives n × (n − 1) × … × (n − r + 1), which equals n! / (n − r)!. This is nPr.

Now suppose order does not matter. Each group of r items has been counted r! times in nPr, once for each way of ordering it. So to count groups, divide by r!. That gives nCr = n! / (r! × (n − r)!). Here n! means n × (n − 1) × … × 1, with 0! = 1.

Two useful consequences follow. First, nCr = nC(n − r), since choosing r people to include is the same as choosing n − r to leave out. Second, nC0 = nCn = 1.

The decision test

Read the problem and ask: if I swap two of the chosen items, do I get a different outcome? Words like select, choose, committee, team, group, hand of cards and handshake signal combinations. Words like arrange, line up, rank, seat, form a number, form a word and assign distinct posts signal permutations. Always confirm with the swap test, because wording alone can mislead.

Worked example 1: same people, two questions

From 5 students, (a) choose a committee of 3, and (b) choose a president, vice-president and secretary.

(a) A committee has no internal roles, so order is irrelevant. Number of ways = 5C3 = 5! / (3! × 2!) = 120 / (6 × 2) = 10.

(b) The three posts are different, so swapping two people changes the outcome. Number of ways = 5P3 = 5! / 2! = 120 / 2 = 60.

Check: 10 × 3! = 10 × 6 = 60. Each committee can fill the three posts in 6 ways, which matches.

Worked example 2: select and then arrange

From 8 people, 3 are chosen and seated in a row of 3 chairs. How many outcomes?

Step 1: choose the 3 people. 8C3 = 8 × 7 × 6 / (3 × 2 × 1) = 336 / 6 = 56.

Step 2: arrange the chosen 3 in the chairs. 3! = 6.

Step 3: multiply, since the two stages happen one after the other. 56 × 6 = 336. This equals 8P3 = 8 × 7 × 6 = 336, as it must.

Worked example 3: "at least" counting

A team of 4 is chosen from 6 boys and 4 girls with at least one girl. Counting directly means adding the cases of 1, 2, 3 and 4 girls. It is faster to count all teams and subtract the ones with no girl.

All teams: 10C4 = 10 × 9 × 8 × 7 / 24 = 5040 / 24 = 210. Teams with no girl, so all 4 are boys: 6C4 = 6C2 = 15. Answer: 210 − 15 = 195.

Mistakes students actually make

• Using the formula before reading for order. Students see "5 people" and reach for 5P3 or 5C3 on reflex. Decide first, calculate second.

• Treating "and" and "or" loosely. Stages that happen together multiply; mutually exclusive cases add. In example 2 we multiplied; in "at least" problems the direct method adds cases.

• Forgetting that a handshake or a diagonal is a pair. Handshakes among 10 people are 10C2 = 45, not 10P2 = 90, because A shaking B's hand is B shaking A's. Diagonals of an n-sided polygon are nC2 − n, since choosing any 2 vertices gives a side or a diagonal. For n = 8 that is 28 − 8 = 20.

• Ignoring identical items. When objects repeat, counting them as distinct overcounts. Dividing by the factorial of each repeat fixes it.

• Dropping the 0! = 1 convention and getting stuck at nCn.

How it appears in JEE

JEE Main usually asks direct numerical questions: forming committees with conditions, counting numbers from given digits, handshakes, diagonals, and cards. JEE Advanced layers restrictions and groupings on top, so a student who has not internalised the swap test loses time on the first line. Expect "at least" and "at most" conditions, words from letters, and selections that are followed by arrangements as in example 2.

Pro Tip: Whenever a problem has both a selection stage and an arrangement stage, do them separately and multiply: nCr × r!. If you find yourself computing nPr and then dividing, or computing nCr and then multiplying, you are using the link nPr = nCr × r! correctly, and it is also the fastest way to check your answer. A teacher at Sukriti NeetJee would ask you to verify every answer this way before moving on.

Frequently Asked Questions

Is a lock combination a combination or a permutation?

Mathematically it is a permutation, because the order of digits matters. Entering 3-1-4 opens it but 4-1-3 does not. The everyday name is misleading, so rely on the swap test and not on the word.

Why is 0! equal to 1?

Because nCn must equal 1: there is exactly one way to choose all n items. The formula gives n! / (n! × 0!), which is 1 only if 0! = 1. It also fits the idea that there is exactly one way to arrange nothing.

When do I add and when do I multiply?

Multiply when one task is followed by another and both must happen. Add when the cases are alternatives that cannot occur together. Ask yourself whether the word "and" or "or" fits between the stages.

Can nCr ever be larger than nPr?

No. Since nPr = nCr × r! and r! is at least 1, nPr is always at least nCr. They are equal only when r is 0 or 1.

Does this chapter matter for NEET?

No, permutations and combinations are part of JEE Mathematics. NEET has no Mathematics paper, although the counting habit helps with probability in genetics problems.

Stuck on a counting problem? Drop it in the comments and we will work through it together, and share this with a fellow JEE aspirant who still guesses between P and C.