## Precalculus (10th Edition)

The combinations are: $\text{abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde}$. $C(5, 3)=10$ combinations.
There are $10$ combinations, namely: $\text{abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde}$ We know that $C(n,r)=\dfrac{n!}{(n-r)!r!}$ Hence, $C(5,3)=\dfrac{5!}{(5-3)!\cdot3!}=\dfrac{5\cdot4\cdot3!}{2\cdot1\cdot3!}=\dfrac{5\cdot4}{2}=10$.