## Precalculus (10th Edition)

There are four combinations are: $123, 124, 134, 234.$ $C(4, 3)=4$
The combinations are: $123, 124, 134, 234.$ We know that $C(n,r)=\dfrac{n!}{(n-r)!r!}$. Hence, $C(4,3)=\dfrac{4!}{(4-3)!\cdot3!}=\dfrac{4\cdot3!}{1\cdot3!}=\dfrac{4}{1}=4$