Question: If $n = 51! +1$, Then find no of primes among $n+1,n+2,\ldots, n+50$

Really speaking, I don't have any clue ...

Answer: 
The number $51!$ has as non-trivial factors every natural number preceding $51$. 

Thus any $51! + 2$, $51! + 3$, etc. will be divisible by $2,3,4,\cdots$ respectively.
$$n=51! + 1$$
$$n+a = 51! + (1+a) = \left(\frac{51!}{1+a}+1\right)\cdot (1+a)$$
If $1\leq a \leq 50$ then $\frac{51!}{1+a}$ is an integer and so $a+1$ divides $n+a$. (and they are obviously not equal, and $a+1\neq 1$). 

Thus there are no primes in the range specified!
The offset by $1$ on $51!$ may make it a little confusing. For any integer $2 \leq n \leq 51$, you will have that $51! + n$ is divisible by $n$ because $51!$ is divisible by $n$. This is an equivalent statement to your problem. Now you just need to show $51!$ is divisible by $n$ and that $n$ does not equal $51! + n$.