If $(1+x)^n = p_0 + p_1 x + p_2 x^2 + \ldots + p_n x^n$,then the value of $p_0 + p_3 + p_6 + \ldots$ is equal to:

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .... + C_nx^n$,then the value of $C_0 + 2C_1 + 3C_2 + .... + (n + 1)C_n$ will be

Let $\binom{n}{k}$ denote ${}^{n}C_{k}$ and $\left[\begin{array}{c} n \\ k \end{array}\right]=\begin{cases} \binom{n}{k}, & \text{if } 0 \leq k \leq n \\ 0, & \text{otherwise} \end{cases}$. If $A_{k}=\sum_{i=0}^{9}\binom{9}{i}\left[\begin{array}{c} 12 \\ 12-k+i \end{array}\right]+\sum_{i=0}^{8}\binom{8}{i}\left[\begin{array}{c} 13 \\ 13-k+i \end{array}\right]$ and $A_{4}-A_{3}=190p$,then $p$ is equal to:

$A$ binary sequence is an array of $0$'s and $1$'s. The number of $n$-digit binary sequences which contain an even number of $0$'s is

If $(1+x)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n$ and $a_0 - a_2 + a_4 - a_6 + \ldots = k \cos \frac{n \pi}{4}$,then $k = $

If $\binom{40}{0} + \binom{41}{1} + \binom{42}{2} + \dots + \binom{60}{20} = \frac{m}{n} \binom{60}{20}$,where $m$ and $n$ are coprime,then $m+n$ is equal to