Let $S = \frac{1}{25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots$ up to $13$ terms. If $13S = \frac{2^{k}}{n!}$ where $k \in N$, then $n + k$ is equal to

If $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + .... + a_{2n}x^{2n}$,then $a_0 + a_2 + a_4 + .... + a_{2n} = $

The value of $-{ }^{15}C_{1} 2 \cdot { }^{15}C_{2} - 3 \cdot { }^{15}C_{3} \ldots - 15 \cdot { }^{15}C_{15} { }^{14}C_{1} { }^{14}C_{3} { }^{14}C_{5} \ldots { }^{14}C_{11}$ is

$^{15}C_3 + ^{15}C_5 + \ldots + ^{15}C_{15} = ?$

If $(\frac{1}{^{15}C_{0}}+\frac{1}{^{15}C_{1}})(\frac{1}{^{15}C_{1}}+\frac{1}{^{15}C_{2}})...(\frac{1}{^{15}C_{12}}+\frac{1}{^{15}C_{13}}) = \frac{a^{13}}{^{14}C_{0} \cdot ^{14}C_{1} \cdot ... \cdot ^{14}C_{12}}$, then $30a$ is equal to:

Let $S_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $S_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
Assertion $(A) : S_3 = 55 \times 2^9$
Reason $(R) : S_1 = 90 \times 2^8$ and $S_2 = 10 \times 2^8$