A permutation counts arrangements where order matters; a combination counts selections where order does not matter. Permutations use nPr = n! ÷ (n−r)! and combinations use nCr = n! ÷ [r! × (n−r)!], so for the same n and r, the permutation count is always r! times larger.
The one-question test settles almost every exam problem: if rearranging the chosen items gives a different outcome, it is a permutation — otherwise, a combination. Electing 3 officers from 10 people is a permutation; choosing any 3 for a committee is a combination. One missing r! is the difference between 720 and 120.
What Is a Permutation?
A permutation is an arrangement of items in a particular order: nPr = n! ÷ (n − r)!. For the first position you have n choices, for the second n − 1, down to (n − r + 1) for the last — the (n − r)! cancels the tail of n! you do not need.
Worked example: 10 people, 3 offices. A club elects a president, a vice-president, and a secretary. Order matters — president is a different job from secretary — so this is a permutation: 10P3 = 10! ÷ (10 − 3)! = 10! ÷ 7! = 10 × 9 × 8 = 720. Real-life permutations: passwords (1234 is not 4321), race rankings, seating arrangements, forming numbers from digits. A free permutation calculator computes nPr with the cancellation shown, so you can check your hand work.
What Is a Combination?
A combination is a selection of items where order does not matter. The number of ways to choose r items from n distinct items is nCr = n! ÷ [r! × (n − r)!] — the permutation formula with one extra division by r!, the number of ways to arrange the r chosen items among themselves.
Worked example: 10 people, a 3-person committee. The same club chooses any 3 members for a committee. Alice, Bob, and Carol are the same committee in any order, so this is a combination: 10C3 = 10! ÷ [3! × 7!] = (10 × 9 × 8) ÷ (3 × 2 × 1) = 720 ÷ 6 = 120 — exactly 3! = 6 times smaller than 720, because each committee can be arranged in 6 orders. Real-life combinations: picking a team, choosing lottery numbers, dealing a poker hand, picking test questions. A free combination calculator runs the same nCr calculation with the working shown.
How Do You Tell Them Apart?
When a problem does not say “arrange” or “choose” outright, ask: if I rearrange the chosen items, do I get a different outcome?
- Yes → permutation. Rearranging the president and secretary changes the leadership; rearranging the digits 1, 2, 3 gives a different number (123 vs 321); rearranging the runners changes who gets gold.
- No → combination. Rearranging the committee members gives the same committee; rearranging the 5 cards in your hand gives the same hand.
Two more cues help. Different roles, ranks, or positions (first/second/third, president/vice-president) signal a permutation; choose, select, pick, or form a group almost always signals a combination. Watch for traps: “choose 3 books to read in order” is a permutation despite the word “choose” — the one-question test catches what keyword-spotting misses.
Permutation vs Combination: Side by Side
| Feature | Permutation (nPr) | Combination (nCr) |
|---|---|---|
| Order matters? | Yes — AB ≠ BA | No — AB = BA |
| What it counts | Arrangements | Selections |
| Formula | n! ÷ (n − r)! | n! ÷ [r! × (n − r)!] |
| Relationship | nPr = nCr × r! | nCr = nPr ÷ r! |
| 10 items, take 3 | 10P3 = 720 | 10C3 = 120 |
| 5 items, take 3 | 5P3 = 60 | 5C3 = 10 |
| 8 items, take 3 | 8P3 = 336 | 8C3 = 56 |
| Keywords | arrange, order, rank, sequence, position | choose, select, committee, team, group |
| Typical examples | Passwords, race podiums, electing officers | Teams, lottery numbers, card hands |
Memorize the relationship row: nPr = nCr × r!. Check it: 10C3 × 3! = 120 × 6 = 720 = 10P3 ✓; 5C3 × 3! = 10 × 6 = 60 = 5P3 ✓; 8C3 × 3! = 56 × 6 = 336 = 8P3 ✓. When r = 0 or r = 1, the two counts are equal (nP0 = nC0 = 1, nP1 = nC1 = n) — otherwise permutations are strictly larger.
The 52-Card Test: One Deck, Two Questions
A standard deck of 52 cards makes the difference concrete — the same deck answers both kinds of question, and the answers are wildly different.
Permutation: how many ways can 5 cards be dealt in order? Order matters, so use nPr: 52P5 = 52! ÷ 47! = 52 × 51 × 50 × 49 × 48 = 311,875,200.
Combination: how many 5-card poker hands exist? A hand is the same hand regardless of the order the cards were dealt, so use nCr: 52C5 = 52! ÷ [5! × 47!] = 311,875,200 ÷ 120 = 2,598,960.
Same deck, same 5 cards — but 311,875,200 ordered deals versus 2,598,960 hands, exactly 5! = 120 times apart. (Note: a common misprint drops a digit — the verified value is 2,598,960.) Both use the same factorial cancellation, the step a factorial calculator shows for n = 52, r = 5 in either mode — and the cancelling is identical to simplifying a fraction, which is why the fraction calculator shows the same divide-top-and-bottom step.
4 Mistakes That Cost Marks
1. Using nCr when order matters. Electing officers is the classic trap: 10C3 = 120 counts committees, but office-holders need 10P3 = 720. Apply the one-question test before choosing a formula.
2. Forgetting to divide by r! in combinations. Computing 10C3 as 10! ÷ 7! = 720 gives the permutation answer. The r! is the entire difference between the two formulas — write it every time.
3. Using nPr when r = n without simplifying. Arranging all n items is just n!: 4P4 = 4! = 24. Similarly, nCn = 1 — one way to choose everything.
4. Entering r > n. Both formulas require r ≤ n — (n − r)! of a negative number is undefined. Check r ≤ n before calculating.
Frequently Asked Questions
What is the difference between permutation and combination?
Order. A permutation counts arrangements where order matters (AB ≠ BA) using nPr = n! ÷ (n−r)!; a combination counts selections where order does not matter (AB = BA) using nCr = n! ÷ [r! × (n−r)!]. For the same n and r, nPr is always r! times larger than nCr.
How do you know if a problem is a permutation or combination?
Ask: if I rearrange the chosen items, do I get a different outcome? If yes, a permutation (ranking, electing officers, passwords); if no, a combination (choosing a team, dealing a hand).
What is 10P3 and 10C3?
10P3 = 10! ÷ 7! = 10 × 9 × 8 = 720, and 10C3 = 10! ÷ (3! × 7!) = 720 ÷ 6 = 120. The 6-fold difference is 3! — each 3-person selection can be arranged in 6 orders.
How many 5-card poker hands are there?
2,598,960 — that is 52C5 = 52! ÷ (5! × 47!) = 311,875,200 ÷ 120. The ordered version, 52P5 = 311,875,200, is 120 times larger because each hand can be dealt in 5! = 120 orders.
Keep Practicing
Work these by hand and verify with the free factorial calculator: 8 runners, top 3 podium — 8P3 = 336; choose 3 from 8 for a team — 8C3 = 56; arrange 4 books — 4P4 = 4! = 24; choose 4 books from 10 — 10C4 = 210; 7 questions, answer any 5 — 7C5 = 21.