$\begin{array}{1 1}\large\frac{n(n+1)(n+2)(2n+3)}{4} \\ \large\frac{n(n+1)(n+2)(n+3)}{2} \\\large\frac{n(n+1)(n+2)(n+3)}{6} \\ \large\frac{n(n+1)(n+2)(n+3)}{4}\end{array} $