If  $\sum\limits_{K = 1}^{12} {12K{.^{12}}{C_K}{.^{11}}{C_{K - 1}}} $ is equal to $\frac{{12 \times 21 \times 19 \times 17 \times ........ \times 3}}{{11!}} \times {2^{12}} \times p$ then $p$ is

Let ${\left( {1 + x} \right)^{10}} = \sum\limits_{r = 0}^{10} {{C_r}{x^r}} $ and ${\left( {1 + x} \right)^7} = \sum\limits_{r = 0}^7 {{d_r}{x^r}} $ . If $P = \sum\limits_{r = 0}^5 {{C_{2r}}} $ and $Q = \sum\limits_{r = 0}^3 {{d_{2r + 1}}} $ , then $\frac{P}{{2Q}}$ is equal to

If ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .... + {C_n}{x^n}$, then ${C_0}{C_2} + {C_1}{C_3} + {C_2}{C_4} + {C_{n - 2}}{C_n}$ equals

If $S_n =$$\sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and $T_n =$$\sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $ then $\frac{{{T_n}}}{{{S_n}}}$ is equal to

$\frac{{{C_0}}}{1} + \frac{{{C_2}}}{3} + \frac{{{C_4}}}{5} + \frac{{{C_6}}}{7} + ....$=

The coefficient of $t^{50}$ in $(1 + t^2)^{25} (1 + t^{25}) (1 + t^{40}) (1 + t^{45}) (1 + t^{47})$ is