The core question: does order matter?
Permutations and combinations both count the number of ways to choose r items from a group of n distinct items, but they answer two different questions. A permutation counts the number of ways to choose and arrange r items, so two selections containing the identical items in a different order are counted as two separate outcomes. A combination counts only the number of ways to choose r items, so two selections containing the identical items are counted as a single outcome regardless of the order they were picked in.
A simple way to decide which one applies: picture two selections that use exactly the same items but list or arrange them differently. If those two selections represent genuinely different real-world results — different rankings, different passwords, different finishing orders in a race — order matters and a permutation is the correct count. If those two selections represent the same real-world result — the same committee, the same hand of cards, the same set of lottery numbers — order does not matter and a combination is the correct count.
The formulas: nPr and nCr, with a verified example
The number of permutations of r items chosen from n distinct items is nPr = n! / (n − r)!, and the number of combinations is nCr = n! / [r! (n − r)!] — the combination formula is the permutation formula divided by r!, because dividing by r! removes the r! different orderings of each same-item group that the permutation count treats as separate.
Worked example: choosing 3 items from a group of 7. Permutations: 7P3 = 7! / (7−3)! = 7 × 6 × 5 = 210. Combinations: 7C3 = 7! / (3! × 4!) = (7 × 6 × 5) / (3 × 2 × 1) = 210 / 6 = 35. The combination count is exactly the permutation count divided by 3! = 6, confirming the relationship nCr = nPr / r!.
| Quantity | Formula | Value for n=7, r=3 |
|---|---|---|
| Permutations (order matters) | nPr = n! / (n − r)! | 7P3 = 7 × 6 × 5 = 210 |
| Combinations (order doesn't matter) | nCr = n! / [r!(n − r)!] | 7C3 = 210 / 3! = 35 |
With repetition vs without repetition
Both formulas above assume each of the n items can be selected at most once — sampling without repetition. When items can be reused (sampling with repetition, also called sampling with replacement), the counting formulas change. Permutations with repetition allowed simply raise n to the power r, since each of the r positions independently has all n choices available regardless of what was chosen before: n^r.
Combinations with repetition allowed use a different formula, C(n + r − 1, r), which counts the number of ways to choose r items from n types when repeats are allowed and order still does not matter — this is a less commonly needed case but arises, for example, when choosing scoops of ice cream from n flavors where the same flavor can be picked more than once.
| Case | Formula | Typical example |
|---|---|---|
| Permutations, no repetition | n! / (n − r)! | Ranking 3 winners from 7 finalists |
| Permutations, with repetition | n^r | A 4-digit PIN where digits can repeat |
| Combinations, no repetition | n! / [r!(n − r)!] | Choosing 6 numbers for a lottery ticket |
| Combinations, with repetition | (n + r − 1)! / [r!(n − 1)!] | Choosing 3 scoops from 5 ice-cream flavors, repeats allowed |
Worked examples: a PIN code and a lottery ticket
A 4-digit PIN, where each of the 4 positions can be any digit 0–9 and digits can repeat (the typical rule for bank PINs), is an example of permutations with repetition: order matters (1234 and 4321 are different PINs) and repetition is allowed, giving 10^4 = 10,000 possible PINs. If repeated digits were not allowed instead, the count would drop to the permutation-without-repetition value, 10P4 = 10 × 9 × 8 × 7 = 5,040 — fewer possibilities, because each digit used removes one option for the remaining positions.
A classic lottery format that draws 6 numbers from a pool of 49, with no number repeated and the drawn order irrelevant to whether a ticket wins, is an example of combinations without repetition: C(49, 6) = 49! / (6! × 43!) = 13,983,816. This large figure is exactly why such lottery formats are described as long-odds games — a combination count grows very large very quickly as n grows relative to r, since it is counting unordered groups out of a large pool rather than a small, fixed set of arrangements.
A quick way to remember which formula to use
"Permutation" and "position" both begin with a P-sound, and a helpful memory aid is that permutations care about position — where in the sequence each item lands. "Combination" describes a combined group, like the contents of a bag or a committee, where only membership matters, not the sequence in which members were added.
Passwords, PINs, race rankings, and any seating or scheduling arrangement are permutation problems, because swapping the order changes the outcome. Card hands, lottery number sets, committee memberships, and any unordered group selection are combination problems, because the members are the only thing that matters, not the sequence in which they were chosen.
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What is the difference between a permutation and a combination?
A permutation counts arrangements where order matters, so choosing the same items in a different order counts as a different outcome. A combination counts selections where order does not matter, so the same items in any order count as a single outcome. Both start from the same group of n items and involve choosing r of them, but a permutation count is always larger than or equal to the corresponding combination count for the same n and r.
What is 7P3 and how do you calculate it?
7P3 equals 210, the number of ways to choose and arrange 3 items from a group of 7 distinct items where order matters. It is calculated as 7P3 = 7! / (7 − 3)! = 7 × 6 × 5 = 210, multiplying the three largest factors of 7! together since the smaller factors cancel with (7−3)! in the denominator.
What is 7C3 and how do you calculate it?
7C3 equals 35, the number of ways to choose 3 items from a group of 7 distinct items where order does not matter. It is calculated as 7C3 = 7! / (3! × 4!) = (7 × 6 × 5) / (3 × 2 × 1) = 210 / 6 = 35 — exactly the permutation count, 7P3 = 210, divided by 3! (the number of ways to reorder each group of 3), since combinations collapse all orderings of the same group into one outcome.
Why do PINs use permutations and lotteries use combinations?
A PIN treats 1234 and 4321 as different codes, so the order of digits changes the outcome, which is a permutation scenario — with repetition allowed for a typical 4-digit PIN, giving 10^4 = 10,000 possibilities. A lottery ticket wins based only on which numbers were drawn, not the order they came out in, so two draws with the same numbers in a different order are the same winning outcome, which is a combination scenario — for a 6-from-49 format, C(49,6) = 13,983,816 possible number sets.
What does 'with repetition' mean in a permutation or combination formula?
It means an item can be selected more than once within the same outcome. Permutations with repetition allowed use n^r (each of r positions independently has all n choices), while combinations with repetition allowed use (n + r − 1)! / [r!(n − 1)!]. Without repetition, each item can be used at most once, which is what the standard nPr and nCr formulas assume.
Kaynaklar
- NIST Digital Library of Mathematical Functions (DLMF), §26 Combinatorial Analysis. dlmf.nist.gov.
- Rosen KH. Discrete Mathematics and Its Applications. 8th ed. McGraw-Hill, 2018 (permutations, combinations, and counting with repetition).
- Ross SM. A First Course in Probability. Pearson (combinatorial counting methods).
- Weisstein EW. "Permutation" and "Combination." MathWorld — A Wolfram Web Resource. mathworld.wolfram.com.