Let S denote the set of 4-digit numbers $a b c d$ such that $a>b>c>d$ and P denote the set of 5 -digit numbers having product of its digits equal to 20 . Then $n(\mathrm{~S})+n(\mathrm{P})$ is equal to $\_\_\_\_$

The number of 4 -letter words, with or without meaning, which can be formed using the letters PQRPQRSTUVP, is $\_\_\_\_$ .

Let ABC be a triangle. Consider four points $\mathrm{p}_1, \mathrm{p}_2, \mathrm{p}_3, \mathrm{p}_4$ on the side AB , five points $p_5, p_6, p_7, p_8, p_9$ on the side $B C$, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC . None of these points is a vertex of the triangle ABC . Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, \ldots, p_{13}$, is $\_\_\_\_$

Let $S=\{(m, n): m, n \in\{1,2,3, \ldots . ., 50\}\}$. If the number of elements $(m, n)$ in $S$ such that $6^m+9^n$ is a multiple of 5 is $p$ and the number of elements ( $m, n$ ) in $S$ such that $m+n$ is a square of a prime number is q , then $\mathrm{p}+\mathrm{q}$ is equal to $\_\_\_\_$ .