## Precalculus (10th Edition)

$1,n$
For $n\choose j$ the correct formula is $\frac{n!}{(n-j)!j!}$. Hence ${n\choose 0}=\frac{n!}{(n-0)!0!}=\frac{n!}{(n)!\cdot1}=1$ , ${n\choose 1}=\frac{n!}{(n-1)!1!}=\frac{n!}{(n-1)!\cdot1}=n$