## Algebra 2 (1st Edition)

We know that $_nC_r=\frac{n!}{r!(n-r)!}$. Hence $_nC_r\cdot_rC_m=\frac{n!}{r!(n-r)!}\frac{r!}{m!(r-m)!}=\frac{n!}{(n-r)!m!(r-m)!}$ $_nC_m\cdot_{n-m}C_{r-m}=\frac{n!}{m!(n-m)!}\frac{(n-m)!}{(r-m)!(n-m-(r-m))!}=\frac{n!}{m!(n-m)!}\frac{(n-m)!}{(r-m)!(n-r)!}=\frac{n!}{(n-r)!m!(r-m)!}$ Thus they are equal; thus we proved what we had to.