high School⏱ 4 min

Combinations C(n,k)

Choosing k from n without order

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Choose 2 from 5 items:

1
2
3
4
5

There are 10 different ways to choose 2 items from 5.

Values

C(5,2)=5!2!⋅3!=1202⋅6=10C(5,2) = \frac{5!}{2! \cdot 3!} = \frac{120}{2 \cdot 6} = 10

C(n,k) counts selections regardless of order. We divide by k! because k items can be arranged k! ways - all representing the same selection.