Option 1:-
we can write product of $n$ consecutive integer as ,
$\prod_{k=1}^{n}(p+k)$
=$(p+1)(p+2)(p+3).....(p+n)$
=$\large \frac{(p+n)!}{p!}$
=$\large \frac{(p+n)! * n!}{p!*n!}$
=$\large \binom{p+n}{p}* n!$
So this term is divisible by $n!$ .
So option (a) is true .
Option 2:-
$\large \sum_{i=0}^{n}\binom{n}{i}$=$\large \binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+......+\binom{n}{n}$
Now we know from binomial theorem ,
$\large (1+x)^{n}=\binom{n}{0}x^{0}+\binom{n}{1}x+\binom{n}{2}x^{2}+\binom{n}{3}x^{3}+......+\binom{n}{n}x^{n}$
So putting x=1 , we get ,
$\large (1+1)^{n}=\binom{n}{0}1^{0}+\binom{n}{1}1+\binom{n}{2}1^{2}+\binom{n}{3}1^{3}+......+\binom{n}{n}1^{n}$
$\large 2^{n}=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+......+\binom{n}{n}$
So ,
$\large \sum_{i=0}^{n}\binom{n}{i}$=$\large 2^{n}$
So option (2) is also true.
Option 3:-
RHS:-
$\large \binom{n-1}{i}+\binom{n-1}{i-1}$
= $\large \frac{(n-1)!}{i!(n-1-i)!}+ \frac{(n-1)!}{(i-1)!(n-i)}$
= $\large (n-1)! \left [ \frac{1}{i!(n-i-1)!}+\frac{1}{(i-1)!(n-i)!}\right ]$
=$\large \frac{(n-1)!}{(i-1)!(n-i-1)!} \left [ \frac{1}{i}+\frac{1}{(n-i)}\right ]$
=$\large \frac{(n-1)!}{(i-1)!(n-i-1)!} \left [ \frac{n}{i(n-i)}\right ]$
=$\large \frac{n!}{i!(n-i)!} =\binom{n}{i}$
So option (C) is true.
Option 4:-
It is nothing but little fermat theorem .
If $n$ is a prime number then for any integer $a$ , then the number $\large a^{n}-a$ is an integer multiple of $n$.
So , Say $a=2$ .$n$ is any odd prime integer.
So by fermat’s theorem we can write,
$\large 2^{n}\equiv 2 mod(n)$
$\large 2^{n-1}\equiv 1 mod(n)$
So this implies ,
$\large 2^{n-1} -1$ divides by n . (This statement is also true) .
Ref:-https://en.wikipedia.org/wiki/Fermat%27s_little_theorem
Option 5:-
counter example:-
It is false as if we take n=3,
then ,$\large \binom{2n}{n}=\binom{6}{3}=20$ it is not divisible by 3 so it is false .
