If ${}^{21}{C_1} + 3.{}^{21}{C_3} + 5.{}^{21}{C_5} + ......19{}^{21}{C_{19}} + 21.{}^{21}{C_{21}} = k$ Then number of prime factors of $k$ is

If ${C_0},{C_1},{C_2},.......,{C_n}$ are the binomial coefficients, then $2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + ....$ equals

If the sum of the coefficients in the expansion of ${(\alpha {x^2} - 2x + 1)^{35}}$ is equal to the sum of the coefficients in the expansion of ${(x - \alpha y)^{35}}$, then $\alpha $=

The sum of the coefficients of three consecutive terms in the binomial expansion of $(1+ x )^{ n +2}$, which are in the ratio $1: 3: 5$, is equal to

Total number of terms in the expansion of $\left[ {{{\left( {1 + x} \right)}^{100}} + {{\left( {1 + {x^2}} \right)}^{100}}{{\left( {1 + {x^3}} \right)}^{100}}} \right]$ is

Let $X =\left({ }^{10} C _1\right)^2+2\left({ }^{10} C _2\right)^2+3\left({ }^{10} C _3\right)^2+\ldots \ldots . .+10\left({ }^{10} C _{10}\right)^2$ where ${ }^{10} C _{ r }, r \in\{1,2, \ldots ., 10\}$ denote binomial coefficients. Then, the value of $\frac{1}{1430} X$ is. . . . . . .