The value of$^n{C_1}\sum\limits_{r = 0}^1 {^1{C_r}} { + ^n}{C_2}\left( {\sum\limits_{r = 0}^2 {^2{C_r}} } \right){ + ^n}{C_3}\left( {\sum\limits_{r = 0}^3 {^3{C_r}} } \right) + ......{ + ^n}{C_n}\left( {\sum\limits_{r = 0}^n {^n{C_r}} } \right)$ is equal to

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

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

The value of $\sum_{ r =0}^{6}\left({ }^{6} C _{ r }{ }^{-6} C _{6- r }\right)$ is equal to :

If ${C_r}$ stands for $^n{C_r}$, the sum of the given series $\frac{{2(n/2)!(n/2)!}}{{n!}}[C_0^2 - 2C_1^2 + 3C_2^2 - ..... + {( - 1)^n}(n + 1)C_n^2]$, Where $n$ is an even positive integer, is

If ${}^{21}{C_1} + 3.{}^{21}{C_3} + 5.{}^{21}{C_5} + ......19{}^{21}{C_{19}} + 21.{}^{21}{C_{21}} = k$ Then number of prime factors of $k$ is