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.
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!