Let $n = 1! + 4! + 7! + \ldots + 400!$. Then the ten's digit of $n$ is

Let the eleven letters $A, B, ....., K$ denote an arbitrary permutation of the integers $(1, 2, ....., 11)$. Then,what is the nature of the product $(A - 1)(B - 2)(C - 3) ..... (K - 11)$?

If ${ }^{1} P_{1}+2 \cdot{ }^{2} P_{2}+3 \cdot{ }^{3} P_{3}+\ldots+15 \cdot{ }^{15} P_{15}={ }^{q} P_{r}-s$,where $0 \leq s \leq 1$,then ${ }^{q+s} C_{r-s}$ is equal to .... .

If the $LCM$ of $p$ and $q$ is $r^2t^4s^2$,where $r, s,$ and $t$ are prime numbers,and $p$ and $q$ are positive integers,then the number of ordered pairs $(p, q)$ is ....

If $a_n = \sum_{r=0}^n \frac{1}{\binom{n}{r}}$,then $\sum_{r=0}^n \frac{r}{\binom{n}{r}} = .....$

The units digit of $\sum\limits_{1 < p < 100} {p! - \sum\limits_{n = 1}^{50} {(2n)!} }$ is: