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 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 $51! + n$.