The coefficient of $x^{4}$ in the expansion of $\left(1+x+x^{2}+x^{3}\right)^{6}$ in powers of $x,$ is

Let $m, n \in N$ and $\operatorname{gcd}(2, n)=1$. If $30\left(\begin{array}{l}30 \\ 0\end{array}\right)+29\left(\begin{array}{l}30 \\ 1\end{array}\right)+\ldots+2\left(\begin{array}{l}30 \\ 28\end{array}\right)+1\left(\begin{array}{l}30 \\ 29\end{array}\right)= n .2^{ m }$ then $n + m$ is equal to (Here $\left.\left(\begin{array}{l} n \\ k \end{array}\right)={ }^{ n } C _{ k }\right)$

If ${a_k} = \frac{1}{{k(k + 1)}},$ for $k = 1,\,2,\,3,\,4,.....,\,n$, then ${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = $

If  $\sum\limits_{K = 1}^{12} {12K{.^{12}}{C_K}{.^{11}}{C_{K - 1}}} $ is equal to $\frac{{12 \times 21 \times 19 \times 17 \times ........ \times 3}}{{11!}} \times {2^{12}} \times p$ then $p$ is

If $1+\left(2+{ }^{49} C _{1}+{ }^{49} C _{2}+\ldots .+{ }^{49} C _{49}\right)\left({ }^{50} C _{2}+{ }^{50} C _{4}+\right.$ $\ldots . .+{ }^{50} C _{ so }$ ) is equal to $2^{ n } . m$, where $m$ is odd, then $n$ $+m$ is equal to.

If $r,k,p \in W,$ then $\sum\limits_{r + k + p = 10} {{}^{30}{C_r} \cdot {}^{20}{C_k} \cdot {}^{10}{C_p}} $ is equal to -