If $n$ be a positive integer such that $n \ge 3$, then the value of the sum to $n$ terms of the series $1 . n - \frac{{\left( {n\, - \,1} \right)}}{{1\,\,!}} (n - 1) + \frac{{\left( {n\, - \,1} \right)\,\,\left( {n\, - \,2} \right)}}{{2\,\,!}} (n - 2) $$-  \frac{{\left( {n\, - \,1} \right)\,\,\left( {n\, - \,2} \right)\,\,\left( {n\, - \,3} \right)}}{{3\,\,!}} (n - 3) + ......$ is 

The coefficient of $x^9$ in the polynomial given by $\sum\limits_{r - 1}^{11} {(x + r)\,(x + r + 1)\,(x + r + 2)...\,(x + r + 9)}$ is

Value $\sum\limits_{r = 0}^{15} {\left( {{}^{15}{C_r}{}^{40}{C_{15}}{}^{20}{C_r} - {}^{35}{C_{15}}{}^{15}{C_r}{}^{25}{C_r}} \right)} $ is-

If $x + y = 1$, then $\sum\limits_{r = 0}^n {{r^2}{\,^n}{C_r}{x^r}{y^{n - r}}} $ equals

If for positive integers $r> 1, n > 2$, the coefficients of the $(3r)^{th}$ and $(r + 2)^{th}$ powers of $x$ in the expansion of $( 1 + x)^{2n}$ are equal, then $n$ is equal to 

Let $K$ be the sum of the coefficients of the odd powers of $x$ in the expansion of $(1+ x )^{99}$. Let a be the middle term in the expansion of $\left(2+\frac{1}{\sqrt{2}}\right)^{200}$. If $\frac{{ }^{200} C _{99} K }{ a }=\frac{2^{\ell} m }{ n }$, where $m$ and $n$ are odd numbers, then the ordered pair $(l, n )$ is equal to :