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: 
First you need to review the definition of the factorial: $$m! = \prod\_{i = 1}^m i = 1 \times 2 \times 3 \times \cdots \times m.$$ This means that $m!$ is divisible by 2, by 3, by 4, by every number up to $m$.

Therefore $51!$ is divisible by 2, by 3, by 4 and by every number up to 51 (and a few others greater than 51, but you don't need to worry about those for this problem).

Then $51! + 2$ is also divisible by 2.

$51! + 3$ is also divisible by 3.

$51! + 4$ is also divisible by 4.

And so on and so forth to $51! + 51$, which is divisible by 51.

Maybe $51! + 1$ is prime. Maybe so is $51! + 53$. But in between those two numbers, there are zero primes.
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.