C(10,3) counts the unordered selections of 3 items chosen from 10. Using C(n,r) = n!/(r!(n-r)!): 10!/(3! × 7!) = 720/6 = 120.

nCr — written C(n,r) or {n}{r} — is the number of ways to choose r items from a set of n distinct items where the order of the selection does not matter. Picking the same three people in a different order is still the same group.

How is a combination different from a permutation?

A combination counts unordered selections; a permutation counts ordered arrangements. Choosing an unordered 3-person committee from 10 people is a combination (120 ways); choosing a president, VP, and secretary from the same 10 people is a permutation (720 ways) because the roles differ.

How does nCr relate to nPr?

nCr = nPr / r!. A permutation counts every ordering of the same r items separately, while a combination treats those r! orderings as one outcome, so dividing by r! converts arrangements into selections. C(10,3) = 720/6 = 120.

Why is C(n,r) equal to C(n,n−r)?

Choosing which r items to include is the same count as choosing which n − r items to leave out — every selection has a unique complement. C(10,3) = C(10,7) = 120.

Why does C(n,0) always equal 1?

Choosing 0 items has exactly one outcome — the empty selection — so C(n,0) = 1 for any n. This matches the formula: n!/(0! × n!) = 1.

When r = n, there is only one way to choose every item — the whole set — so C(n,n) = 1 for any n.

Combinations Calculator (nCr)

Calculate C(n,r), the number of unordered ways to choose r items from n, using nCr = n!/(r!(n−r)!), the nPr/r! relation, and the symmetry rule.

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How to use this calculator

Enter n (the total number of distinct items available) and r (how many you are choosing). The calculator returns C(n, r) — the number of unordered selections — along with the factorial formula, the nPr/r! relation, the symmetry rule, and a full step-by-step breakdown.

Tip: Combinations answer "how many groups of r can I choose from n?" If the order of the outcome matters instead, you want a permutation (nPr) — see the comparison below.

Order doesn't matter — the defining idea

A combination counts unordered selections. Picture a committee of 3 people chosen from a pool of 10 — nobody has a distinct role, so a committee of Alice, Bob, and Carol is the same outcome no matter which order you picked them in. That insensitivity to order is exactly what $nCr$ measures.

C(n,r) = \binom{n}{r} = \frac{n!}{r!\,(n-r)!}

Worked example — C(10,3)

C(10,3) = \frac{10!}{3! \times 7!} = \frac{720}{6} = 120

There are 120 different 3-person committees you can form from 10 people.

The nCr = nPr / r! relation

Every combination corresponds to $r!$ permutations — one for each way to order the same group of items. Dividing the permutation count by $r!$ collapses those orderings back into a single unordered selection:

C(n,r) = \frac{P(n,r)}{r!}

For $P(10,3) = 720$ and $r! = 3! = 6$ : $C(10,3) = 720/6 = 120$ — matching the factorial formula above. This is why the calculator's substitution line always shows nPr divided by r! alongside the factorial ratio.

Symmetry: C(n,r) = C(n,n−r)

Choosing which r items to keep is exactly as hard as choosing which $n-r$ items to leave out — the two choices determine each other. So $C(n,r)$ and $C(n,n-r)$ always give the same count:

C(10,3) = C(10,7) = 120

Picking a 3-person committee from 10 people is the same count as picking the 7 people who aren't on the committee. This symmetry is also why Pascal's triangle is mirror-symmetric across each row — each row's entries are exactly the $nCr$ values for that $n$ .

Combinations vs. permutations

Both start from the same $n$ items, but they answer different questions:

	Combination `nCr`	Permutation `nPr`
Question	In how many ways can I select r items?	In how many orders can I arrange r items?
Formula	$n!/(r!(n-r))!$	$n!/(n-r)!$
Example (10 people, r = 3)	3-person committee: 120 ways	President/VP/Secretary: 720 ways
Relationship	$nCr = nPr / r!$	—

For a poker example: dealing 5 cards from a 52-card deck as an unordered hand gives $C(52,5) = 2{,}598{,}960$ possible hands — a number every poker probability calculation starts from.

Special cases

C(n, 0) = 1 — there is exactly one way to choose zero items: the empty selection.
C(n, n) = 1 — there is exactly one way to choose every item: the whole set.
C(n, 1) = n — choosing a single item from n is just picking which one.
C(n, r) = C(n, n−r) — symmetry: choosing r items to keep is the same count as choosing n − r items to discard.

Frequently asked questions

What is C(10,3)?: C(10,3) counts the unordered selections of 3 items chosen from 10. Using $C(n,r) = n!/(r!(n-r)!)$ : $10!/(3! \times 7!) = 720/6 = 120$ .
What does nCr mean?: nCr — written $C(n,r)$ or $\binom{n}{r}$ — is the number of ways to choose $r$ items from a set of $n$ distinct items where the order of the selection does not matter. Picking the same three people in a different order is still the same group.
How is a combination different from a permutation?: A combination counts unordered selections; a permutation counts ordered arrangements. Choosing an unordered 3-person committee from 10 people is a combination (120 ways); choosing a president, VP, and secretary from the same 10 people is a permutation (720 ways) because the roles differ.
How does nCr relate to nPr?: $nCr = nPr / r!$ . A permutation counts every ordering of the same $r$ items separately, while a combination treats those $r!$ orderings as one outcome, so dividing by $r!$ converts arrangements into selections. $C(10,3) = 720/6 = 120$ .
Why is C(n,r) equal to C(n,n−r)?: Choosing which r items to include is the same count as choosing which n − r items to leave out — every selection has a unique complement. $C(10,3) = C(10,7) = 120$ .
Why does C(n,0) always equal 1?: Choosing 0 items has exactly one outcome — the empty selection — so $C(n,0) = 1$ for any $n$ . This matches the formula: $n!/(0! \times n!) = 1$ .
What is C(n,n)?: When $r = n$ , there is only one way to choose every item — the whole set — so $C(n,n) = 1$ for any $n$ .