$\sum_{r=0}^{10} {}^{40-r} C_5$ is equal to

The value of ${ }^{10} C_{1}+{ }^{10} C_{2}+{ }^{10} C_{3}+\ldots+{ }^{10} C_{9}$ is

If $\frac{{}^{11}C_1}{2} + \frac{{}^{11}C_2}{3} + \dots + \frac{{}^{11}C_9}{10} = \frac{n}{m}$ with $\gcd(n, m) = 1$,then $n + m$ is equal to

If $\sum_{r=0}^5 \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{m}{n}$,$\text{gcd}(m, n) = 1$,then $m - n$ 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 $S_2 = 10 \times 2^8$

Let ${ }^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1+ x )^{ n }.$ If $\sum_{ k =0}^{10}\left(2^{2}+3 k \right){ }^{10} C _{ k }=\alpha \cdot 3^{10}+\beta \cdot 2^{10},$ where $\alpha, \beta \in R,$ then $\alpha+\beta$ is equal to ....... .