If $^{2017}C_0 + ^{2017}C_1 + ^{2017}C_2 + ...... + ^{2017}C_{1008} = \lambda^2$ where $\lambda > 0$,then the remainder when $\lambda$ is divided by $33$ is:

Evaluate the sum: $\left( \binom{21}{1} - \binom{10}{1} \right) + \left( \binom{21}{2} - \binom{10}{2} \right) + \left( \binom{21}{3} - \binom{10}{3} \right) + \dots + \left( \binom{21}{10} - \binom{10}{10} \right) = $

The sum $\sum\limits_{i = 0}^m {\binom{10}{i}} {\binom{20}{m - i}}$,(where $\binom{p}{q} = 0$ if $p < q$),is maximum when $m$ is

$^nC_0 - \frac{1}{2} ^nC_1 + \frac{1}{3} ^nC_2 - \dots + (-1)^n \frac{^nC_n}{n+1} = $

If $(1 + x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$,then $a_0 + a_3 + a_6 + \dots =$

Let $n \in N$ and $[x]$ denote the greatest integer less than or equal to $x$. If the sum of $(n+1)$ terms ${}^{n}C_{0}, 3 \cdot {}^{n}C_{1}, 5 \cdot {}^{n}C_{2}, 7 \cdot {}^{n}C_{3}, \ldots$ is equal to $2^{100} \cdot 101$,then $2\left[\frac{n-1}{2}\right]$ is equal to $....$