#### Find the value of  $\sum_{p=1}^{n}\left(\sum_{m=p}^{n}{ }^{n} C_{m} \cdot{ }^{m} C_{p}\right).$Option: 1 Option: 2 Option: 3 Option: 4

Let $S=\sum_{p=1}^{n}\left(\sum_{m=p}^{n}{ }^{n} C_{m} \cdot{ }^{m} C_{p}\right)$

$=\sum_{p=1}^{n}\left({ }^{n} C_{p} \cdot{ }^{p} C_{p}+{ }^{n} C_{p+1} \cdot{ }^{p+1} C_{p}+\ldots \ldots \ldots+{ }^{n} C_{n} \cdot{ }^{n} C_{p}\right) \\$$=\sum_{p=1}^{n} \text { [coefficient of } x^{p} \text { in }\left\{{ }^{n} C_{p}(1+x)^{p}+{ }^{n} C_{p+1}(1+x)^{p+1}+\ldots \ldots . .+{ }^{n} C_{n}(1+x)^{n}\right. \text { ] } \\$$=\sum_{p=1}^{n} \text { [coefficient of } x^{p} \text { in }\left\{{ }^{n} C_{0}+{ }^{n} C_{1}(1+x)+{ }^{n} C_{2}(1+x)^{2}+\ldots \ldots . .\right. \\$

$\left.\left.+{ }^{n} C_{p-1}(1+x)^{p-1}+{ }^{n} C_{p}(1+x)^{p}+\ldots \ldots \ldots .+{ }^{n} C_{n}(1+x)^{n}\right\}\right] \\$

$=\sum_{p=1}^{n} [co-efficient$ of $x^{p}$ in  $\left.(2+x)^{n}\right]$

$=\sum_{p=1}^{n}\left[{ }^{n} C_{p} \cdot 2^{n-p}\right]$

$={ }^{\mathrm{n}} \mathrm{C}_{1} \cdot 2^{\mathrm{n}-1}+{ }^{\mathrm{n}} \mathrm{C}_{2} \cdot 2^{\mathrm{n}-2}+\ldots \ldots \ldots \ldots \ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}$

$={ }^{n} C_{0} \cdot 2^{n}+{ }^{n} C_{1} \cdot 2^{n-1}+{ }^{n} C_{2} \cdot 2^{n-2}+\ldots \ldots \ldots+{ }^{n} C_{n}-{ }^{n} C_{0} \cdot 2^{n}$$=(1+2)^{n}-2^{n}=3^{n}-2^{n}$