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

$C(5, 3)=10$ The possible combinations are: $abc,abd,abe,acd,ace,ade\\ bcd,bce,bde\\cde$
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, $C(5, 3)=\dfrac{5!}{(5-3)!3!}=\dfrac{5!}{2!3!}=10$ The list of all combinationsL $abc,abd,abe,acd,ace,ade\\bcd,bce,bde\\cde$