## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$C(6, 3)=20$ The posible combinations are: $123,124,125,126,134,135,136,145,146,156\\ 234,235,236,245,246,256\\ 345,346,356\\ 456$
If we want to choose $r$ elements out of $n$ disregarding the order, not allowing repetition, we can do this in $C(n, r)=\frac{n!}{(n-r)!r!}$ ways. Hence, $C(6, 3)=\dfrac{6!}{(6-3)!3!}=\dfrac{6!}{3!3!}=20$ The list of all possible combinations: $123,124,125,126,134,135,136,145,146,156\\ 234,235,236,245,246,256\\ 345,346,356\\ 456$