Permutation vs Combination — Deciding nPr or nCr in Five Seconds
Permutations and combinations are not hard to compute. The hard part is a single decision made in the first five seconds of reading the question: does the order of the chosen items matter? Get that right and the arithmetic is a two-line job. Get it wrong and every number after it is wrong, however neat the working looks.
The two formulas
That third line is the whole idea in one equation. First choose your r items (that is nCr), then arrange them among themselves in r! ways. A permutation is a combination that has been arranged. So nPr is always the bigger number — if your "arrangement" answer comes out smaller than your "selection" answer, you have swapped the two formulas.
What the letters mean
- n — how many distinct things you are choosing from.
- r — how many you actually take. Always r ≤ n for these formulas.
- n! — n factorial, i.e. n × (n−1) × … × 2 × 1. By definition 0! = 1, which is what makes nCn = 1 come out correctly.
- Both results are plain counts, so they carry no unit and are always whole numbers. A fractional answer means an arithmetic slip.
The five-second test
Ask one question: "If I swap two of the chosen items around, have I got something different?"
| Wording in the question | Swap changes the outcome? | Use |
|---|---|---|
| arrange, order, sequence, rank, line up, seat | Yes | nPr |
| president / captain / different prizes / distinct posts | Yes — the roles differ | nPr |
| PIN, password, number formed from digits | Yes — 123 ≠ 321 | nPr |
| select, choose, committee, team, group, sample | No | nCr |
| handshakes, chords, diagonals, lines through points | No — the pair is the same pair | nCr |
| identical prizes, same-value items | No | nCr |
Worked example 1 — the same ten students, two different questions
(a) From 10 students, a committee of 3 is to be formed. In how many ways?
Swapping two committee members changes nothing — it is the same committee. Order does not matter, so use nCr:
10C3 = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120 ways
(b) From the same 10 students, a president, a secretary and a treasurer are to be chosen. In how many ways?
Now swapping two of them does change the outcome — a different person is president. Order matters, so use nPr:
10P3 = 10 × 9 × 8 = 720 ways
The relationship, confirmed: 720 = 120 × 3! = 120 × 6 ✔ Each of the 120 committees can hand out the three posts in 6 different ways.
Worked example 2 — the letters of CHEMISTRY
CHEMISTRY has 9 letters, all different — check this before doing anything else, because repeated letters change the method entirely.
(a) Arrange all 9 letters:
9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362,880 arrangements
(b) Choose 4 letters and arrange them into a 4-letter code:
9P4 = 9 × 8 × 7 × 6 = 3,024
(working: 9 × 8 = 72, 72 × 7 = 504, 504 × 6 = 3,024 codes)
(c) Just choose 4 letters, with no ordering:
9C4 = 3,024 / 4! = 3,024 / 24 = 126 selections
Same n, same r, and the answers differ by a factor of 24 — which is exactly 4!. The wording of part (b) versus part (c) is the only thing that separates 3,024 from 126.
Worked example 3 — team plus captain (both formulas in one question)
Question: From 15 players, a team of 11 is selected and one of the eleven is made captain. In how many ways can this be done?
Step 1 — select the team. A team is a team regardless of order, so this is a combination. Use the symmetry nCr = nCn−r to make the arithmetic easy:
15C11 = 15C4 = (15 × 14 × 13 × 12) / (4 × 3 × 2 × 1)
Numerator: 15 × 14 = 210; 210 × 13 = 2,730; 2,730 × 12 = 32,760
Denominator: 24
15C4 = 32,760 / 24 = 1,365 teams
Step 2 — appoint the captain. Any one of the 11 selected players: 11 ways.
Step 3 — multiply (the two choices are made one after the other, so the
fundamental counting principle applies):
1,365 × 11 = 15,015 ways
Note that choosing 11 out of 15 and choosing the 4 to leave out are the same act — that is what 15C11 = 15C4 means, and it turns a horrible calculation into an easy one.
Worked example 4 — handshakes versus gifts
(a) Twelve people at a meeting shake hands once with every other person. How many handshakes?
A handshake between A and B is the same event as one between B and A, so order does not matter:
12C2 = (12 × 11) / (2 × 1) = 132 / 2 = 66 handshakes
(b) Instead, each person sends a greeting card to every other person. How many cards?
A card from A to B is not the same as a card from B to A — sender and receiver are distinct roles, so order matters:
12P2 = 12 × 11 = 132 cards
Exactly twice the handshake count, because 2! = 2. The same twelve people and the same "every pair", but a different question.
Two cases where neither formula applies
Both nPr and nCr assume distinct items chosen without repetition. Two very common exam situations break that assumption:
- Repetition allowed. A 4-digit PIN using digits 0–9 where digits may repeat is 10 × 10 × 10 × 10 = 104 = 10,000, not 10P4. The rule is nr.
- Identical items in the pool. To arrange the letters of BALLOON — 7 letters with L twice and O twice — divide by the factorial of each repeat count: 7! / (2! × 2!) = 5,040 / 4 = 1,260 arrangements, not 7! = 5,040.
Common mistakes
- Using nPr for a committee. The single most frequent error. "Select a group" never uses a permutation unless the members are given different roles.
- Dividing by r! twice. If you wrote nCr as n!/[r!(n−r)!] you have already divided; do not divide again "because order does not matter".
- Adding when you should multiply. Sequential choices multiply ("and"); mutually exclusive alternatives add ("or"). Team and captain in example 3 was a multiplication.
- Ignoring repeated letters. Always count the letters and look for duplicates before writing n!. BALLOON is a trap; CHEMISTRY is not.
- Cancelling factorials carelessly. 9!/5! = 9 × 8 × 7 × 6, not 9!/5 and not 4!. Write out the cancellation on the page.
- Forgetting 0! = 1. It is a definition, not a special case, and it is what makes nC0 = nCn = 1 work.
Where this appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 11 | Permutations and combinations chapter: word arrangements, committee selections, digit problems |
| CBSE/ICSE Class 12 | Probability — the favourable and total outcome counts are almost always nCr |
| JEE Main & Advanced | Binomial theorem (the coefficients are nCr), distribution and arrangement problems |
| Chemistry connection | Counting microstates in statistical thermodynamics, and the number of ways electrons occupy degenerate orbitals |
Decide first, then check the number. The nPr / nCr calculator gives both values for the same n and r side by side, so you can see the factor of r! between them and confirm that you picked the right one for the wording.
Open the nPr / nCr Calculator →Working towards Class 11–12 boards and competitive papers? ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India, plus home tuition across Delhi-NCR — see abcchemistry.in.