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!
In addition to Michael Hardy's answer, which precludes all of n+1 to n+50, you can also prove that n is in fact composite as well using Wilson's Theorem which states that a number n is prime if and only if (n - 1)! ≡ (n-1) mod n. 

Here, n is (51! + 1), so (n-1)! = 51!!. Therefore, (51! + 1) is prime iff 51!! ≡ 51! mod (51! + 1). However, we now that (51!+1) is of course a factor of 51!! since (51! + 1) < 51!!, so it is impossible that 51!! would be congruent to anything other than 0 modulo (51! + 1). Therefore, 51! + 1 is also definitely not prime.