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  

There are many recurrence relations for $\dbinom{n}{r}$. One of the most commonly used one is the following. $$\binom{n+1}{r} = \binom{n}{r} + \binom{n}{r-1}.$$ There are many ways to prove this and one simple way is to look at $\displaystyle (1+x)^{n+1}$. We know that $$(1+x)^{n+1} = (1+x)^{n} (1+x).$$ Now compare the coefficient of $x^r$ on both sides.

 The left hand the coefficient of $x^r$ is $$\dbinom{n+1}{r}.$$ On the right hand side, the $x^r$ term is obtained by multiplying the $x^r$ term of $(1+x)^n$ with the $1$ from $(1+x)$ and also by multiplying the $x^{r-1}$ term of $(1+x)^n$ with the $x$ from $(1+x)$. Hence, the coefficient of $x^r$ is given by the sum of these two terms from the right hand side i.e. $$\dbinom{n}{r} + \dbinom{n}{r-1}$$
  As Rob and J.M have pointed out, these recurrences define the celebrated Pascal's triangle. A pictorial representation is shown above. Each row represents the value of $n$ starting from $0$. On a row, as we proceed from left to right, the different values of $r$ from $0$ to $n$ are hit. The arrows indicate how the $r^{th}$ value on the $(n+1)^{th}$  row is obtained as the sum of the $(r-1)^{th}$ and $r^{th}$ value on the $n^{th}$ row.

