How this instrument works
A combination is an unordered selection: pick r items out of a pool of n, and any two selections containing the same items count as the same outcome, no matter what order they were picked in. That's the defining test — if picking Alice-then-Bob and Bob-then-Alice should count as one result, not two, you have a combination problem, and the count is written nCr, read "n choose r."
The formula, nCr = n! / (r! × (n − r)!), builds on the ordered count first and then removes the ordering. There are n! / (n − r)! ways to select and arrange r items in order (that's a permutation), but each unordered group of r items can itself be arranged r! different ways, so every group gets counted r! times too many. Dividing the ordered count by r! removes that overcounting, leaving exactly the number of distinct groups — nCr.
This calculator assumes each of the n items can be selected at most once, which is the standard default for the word "combination" — you can't pick the same raffle entry twice. If your situation genuinely allows repeats, like choosing 3 scoops of ice cream from 5 flavors where the same flavor can be picked more than once, the count follows a different formula; this site's combinations-with-repetition instrument handles that case directly.
- Enter the total number of items available into n (total items) — everything eligible to be chosen from, such as raffle entries or candidates.
- Enter how many are actually being selected into r (items chosen).
- Read nCr — combinations directly beneath the inputs; the instrument computes n! / (r! × (n − r)!) for you.
- If r is larger than n, the instrument flags it — you cannot choose more items than the pool actually contains.
- If order matters for your problem instead — say, ranking first, second and third place rather than just picking a group of winners — use this site's permutation instrument, which counts nPr instead.
Worked example — choosing 3 winners from 10 raffle entries
A raffle has 10 entries, and 3 winners are drawn with no distinction between them — just three names pulled, not ranked first, second and third. Enter n = 10 and r = 3. The instrument computes 10! divided by (3! × 7!), which works out to (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120. nCr reads 120.
That 120 counts every distinct group of 3 winners possible from the 10 entries, treating {Alice, Bob, Carol} as the same outcome regardless of which order they were drawn in. If the raffle instead awarded first, second and third prize separately — so the drawing order mattered — the count would be the permutation total instead, 720, exactly 3! = 6 times larger, since each group of 3 winners could be ordered across the three prizes in 6 different ways.
Questions
What does nCr actually count?
The number of distinct unordered groups of r items you can choose from a pool of n, where the order of selection doesn't matter. For n = 10 and r = 3, nCr = 120 — every distinct group of 3 winners possible from 10 raffle entries, regardless of which order they were drawn in.
What's the difference between a combination and a permutation?
A combination ignores order and counts only which items were selected; a permutation counts order as significant. Picking an unordered group of 3 raffle winners from 10 entries is a combination (120 outcomes); awarding first, second and third prize to those same entries is a permutation (720 outcomes) — 6 times larger, since each group of 3 can be ordered 3! = 6 ways.
Why does the formula divide by r! as well as (n − r)!?
Dividing by (n − r)! alone gives the ordered count — the number of ways to pick and arrange r items in sequence. But each unordered group of r items can itself be arranged r! different ways, so that ordered count includes every group r! times over. Dividing by r! again removes that duplication, leaving the count of distinct unordered groups.
Why does choosing all n of n items give exactly 1?
Because there's only one possible group when you select every item in the pool — there's no choice left to make. Formally, nCn = n! / (n! × 0!) = n! / n! = 1, since 0! is defined as 1. Choosing all 5 of 5 items, for example, gives C(5,5) = 1.
Why does choosing 0 items also give exactly 1?
Because there is exactly one way to select nothing: the empty group. Formally, nC0 = n! / (0! × n!) = 1, using 0! = 1. Choosing 0 of 6 items, for example, gives C(6,0) = 1 — it looks like an edge case, but it's a real and standard result.
Does this instrument assume repeats are allowed or not?
It assumes each item can be selected at most once, which is the standard meaning of "combination." If your scenario allows the same item to be chosen more than once — like 3 ice cream scoops from 5 flavors, where a flavor can repeat — use this site's combinations-with-repetition instrument instead, which follows a different formula, C(n + r − 1, r).