## College Algebra (6th Edition)

We know that ${n\choose r}=\frac{n!}{(n-r)!r!}$ by definition Hence here: ${n\choose r}=\frac{n!}{(n-r)!r!}=\frac{n!}{r!(n-r)!}=\frac{n!}{(n-(n-r))!(n-r)!}={n\choose n-r}$. Thus we proved what we wanted to.