If the coefficients of the $5^{th}$,$6^{th}$,and $7^{th}$ terms in the expansion of $(1 + x)^n$ are in $A.P.$,then $n =$

Let $(1 + x)^{10} = \sum_{r=0}^{10} C_r x^r$ and $(1 + x)^7 = \sum_{r=0}^7 d_r x^r$. If $P = \sum_{r=0}^5 C_{2r}$ and $Q = \sum_{r=0}^3 d_{2r+1}$,then $\frac{P}{2Q}$ is equal to

The sum of the series $\frac{1}{1 \times 2} {}^{25}C_{0} + \frac{1}{2 \times 3} {}^{25}C_{1} + \frac{1}{3 \times 4} {}^{25}C_{2} + \ldots + \frac{1}{26 \times 27} {}^{25}C_{25}$ is

If $n$ is a positive integer,then $\sum_{r=1}^n r^2 \cdot C_r = (\ldots \ldots \ldots) 2^{n-2}$

If $A = \left\{ \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} : a_i, b_i, c_i \in \{ \text{binomial coefficients in the expansion of } (1+x)^{11} \} \right\}$,then the number of elements in set $A$ is: (in $^9$)

$\binom{47}{4} + \sum_{r=1}^5 \binom{52-r}{3} = \dots$