Combinations (nCr)

Also known as nCr · how many ways to choose

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

nCr =

n =

r =

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

A combination counts groups, not orders. Start from the permutations n!/(n−r)! and divide by r!, because each group of r items was counted once for every way of shuffling it. A five-card poker hand from a 52-card deck: 52C5 = 2,598,960. A 6/49 lottery ticket: 49C6 = 13,983,816, which is exactly why the jackpot chance is about 1 in 14 million.

These numbers are the entries of Pascal's triangle, which Pascal himself laid out in the Traité du triangle arithmétique (written 1654, printed 1665) while corresponding with Fermat about how to divide the stakes of an interrupted game — the exchange usually credited with founding probability theory. The triangle was known centuries earlier to Halayudha in India and to Yang Hui in China. Two things to remember: nCr = nC(n−r), so choosing 47 cards to discard is the same job as choosing 5 to keep, and the count has no closed-form inverse, so this calculator solves only for the number of groups.

Combinations (nCr)

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

Where

${}_{n}C_{r}$ = Number of unordered selections
$n$ = Items available
$r$ = Items chosen

Missing one of these? Work it out first, then come back

Number of unordered selections — Permutations (nPr), Binomial Coefficient (n Choose k)
Items available — Binomial Coefficient (n Choose k), Permutations (nPr)
Items chosen — Binomial Coefficient (n Choose k), Permutations (nPr)