$\sum_{k=0}^{20} \left({}^{20}C_{k}\right)^{2}$ is equal to :

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .... + C_nx^n$,then $C_0C_2 + C_1C_3 + C_2C_4 + .... + C_{n-2}C_n$ equals

$^{4n}C_0 + ^{4n}C_4 + ^{4n}C_8 + ... + ^{4n}C_{4n}$ is

If $\frac{1}{n+1} {}^{n}C_{n} + \frac{1}{n} {}^{n}C_{n-1} + \dots + \frac{1}{2} {}^{n}C_{1} + {}^{n}C_{0} = \frac{1023}{10}$,then $n$ is equal to

If ${}^{21}C_1 + 3 \cdot {}^{21}C_3 + 5 \cdot {}^{21}C_5 + \dots + 19 \cdot {}^{21}C_{19} + 21 \cdot {}^{21}C_{21} = k$,then the number of prime factors of $k$ is:

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