## Precalculus (10th Edition)

The combinations: $123,124,125,126,134,135,136,145,146,156$ $234,235,236,245,246,256,345,346,356,456$ $C(6,3)=20$
The combinations: $123,124,125,126,134,135,136,145,146,156$ $234,235,236,245,246,256,345,346,356,456$ We know that $C(n,r)=\frac{n(n-1)(n-2)...(n-k+1)}{r!}$. Also $C(n,0)=1$ by convention. Also $C(n,r)=C(n,n-r).$ Hence $C(6,3)=\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1}=20$.