What is  the method to compute $\binom{n}{r}$ in a recursive manner? How do you solve this? 

Find out which recurrence relation involving $\dbinom{n}{r}$ is valid, and thus prove that we can compute $\dbinom{n}{r}$ in a recursive manner.

I appreciate any help. Thank You  

To prove:
$$\dbinom{n}{k}=\dbinom{n-1}{k-1}+\dbinom{n-1}{k}$$
We have:
$$\dbinom{n-1}{k-1}=\frac{(n-1)!}{(k-1)!(n-k)!}=\frac{(n-1)!}{(k-1)!(n-k-1)!}\frac{1}{n-k}$$
$$\dbinom{n-1}{k}=\frac{(n-1)!}{(k)!(n-k-1)!}=\frac{(n-1)!}{(k-1)!(n-k-1)!}\frac{1}{k}$$
Adding the two together:
$$\dbinom{n-1}{k-1}+\dbinom{n-1}{k}=\frac{(n-1)!}{(k-1)!(n-k-1)!}(\frac{1}{n-k}+\frac{1}{k})$$
$$=\frac{(n-1)!}{(k-1)!(n-k-1)!}\frac{n}{k(n-k)}$$
$$=\frac{n!}{k!(n-k)!}$$
$$=\dbinom{n}{k}$$
