Solving $f\left( n \right) = \sum\limits_{i > j \ge 0} {{}^{n + 1}{C_i}{}^n{C_j}} $ 
My approah is as follow
Let $f\left( n \right) = \sum\limits_{i > j \ge 0} {{}^{n + 1}{C_i}{}^n{C_j}} $
$f\left( n \right) = \sum\limits_{i = i}^n {{}^{n + 1}{C_i}} \sum\limits_{j = 0}^i {{}^i{C_j}}  \Rightarrow f\left( n \right) = \sum\limits_{i = i}^{n + 1} {{2^i}.{}^{n + 1}{C_i}} $
${\left( {1 + 2x} \right)^{n + 1}} = {}^{n + 1}{C_0} + {}^{n + 1}{C_1}\left( {2x} \right) + ... + {}^{n + 1}{C_{n + 1}}{\left( {2x} \right)^{n + 1}} = {3^{n + 1}}\left\{ {x = 1} \right\}$
Not able to proceed

The summand is the coefficient of $x^{i-j}$ from the following expression:
$$
(1+x)^{n+1}\left(1+\frac{1}{x}\right)^{n}
=
\frac{(1+x)^{2n+1}}{x^{n}}
$$
Since $i-j>1$, we just need to sum the coefficients of $x^{n+1},...,x^{2n+1}$ from the numerator on RHS.
$$
\binom{2n+1}{n+1}+\binom{2n+1}{n+2}+...+\binom{2n+1}{2n+1}=2^{2n}
$$
Seems like all the options are correct except (B)
