show $nCk= \frac{n}{n-k}\cdot(n-1)C(k) $

Question

It's a formula getting from a recursion programming but i cannot show that $$nCk= \frac{n}{n-k}\cdot(n-1)C(k) $$

Answer 1

Note that

$$nCk=\frac{n!}{(n-k)!k!}=\frac{n\cdot (n-1)!}{(n-k)(n-k-1)!k!}=\frac{n}{n-k}\cdot\frac{(n-1)!}{[(n-1)-k]!k!}$$

Now, what does that last fraction equal?

Answer 2

You can ask yourself: In how many ways can we choose $k+1$ people, including one president, from a population of $n$?

First answer: You choose $k$ people from $n$ people and then from the rest of them a president, so we have ${n\choose k}\cdot {n-k\choose 1}$.

Second answer: We can first choose a president among all $n$ people and then $k$ people among $n-1$ people, so that is ${n\choose 1} \cdot {n-1\choose k}$.

So we have:$${n\choose k}\cdot (n-k)=n{n-1\choose k} $$