Prove ${n \choose k} = {n \choose k-1}\frac{n-k+1}{k}$

Question

I am looking to prove, by induction, the following equality:$${n \choose k} = {n \choose k-1}\frac{n-k+1}{k}$$ From Pascal's identity, I know we have that $${n+1 \choose k} = {n \choose k} + {n \choose k-1}$$ The problem gives the following hint: get ${n \choose k-1}$ and ${n+1 \choose k-1}$ in terms of ${n \choose k-1}$. I have been rearranging Pascal's identity for hours now, and can't seem to come up with a solution.

Thanks!

Answer 1

$$\binom{n}{k}=\frac{n!}{(n-k)!k!}=\frac{(n-k+1)n!}{(n-k+1)!k!}=\frac{n!(n-k+1)}{(n-k+1)!(k-1)!k}=\frac{n!}{(n-k+1)!(k-1)!}\frac{(n-k+1)}{k}=\binom{n}{k-1}\frac{n-k+1}{k}$$

Answer 2

Using your method, $$\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$$ Use $\binom{n+1}{k}=\frac{n+1}{k}\binom{n}{k-1}$ $$\frac{n+1}{k}\binom{n}{k-1}= \binom{n}{k}+\binom{n}{k-1}$$ $$\binom{n}{k}=\left(\frac{n+1}{k}-1\right)\binom{n}{k-1}=\frac{n-k+1}{k}\binom{n}{k-1}$$