If $n = ^mC_2$,then the value of $^nC_2$ is given by

If $x={ }^{16} C_5+{ }^{12} C_4, y=\sum_{r=1}^3{ }^{(20-r)} C_4, z=\sum_{k=1}^4{ }^{(16-k)} C_3$,then $x+y+z=$

The value of ${2^n} \{ 1 \cdot 3 \cdot 5 \cdots (2n - 3) \cdot (2n - 1) \}$ is

All five-letter words are made using all the letters $A, B, C, D, E$ and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_{n}$. Let the probability $P(W_{n})$ of choosing the word $W_{n}$ satisfy $P(W_{n}) = 2P(W_{n-1}), n > 1$. If $P(CDBEA) = \frac{2^{\alpha}}{2^{\beta}-1}, \alpha, \beta \in N$,then $\alpha + \beta$ is equal to

Among the $4$-digit numbers that can be formed using the digits $1, 2, 3, 4, 5, 6$ without repeating any digit,the number of numbers which are divisible by $6$ is:

Suppose $a_2, a_3, a_4, a_5, a_6, a_7$ are integers such that $\frac{5}{7} = \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} + \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!}$,where $0 \leq a_j < j$ for $j = 2, 3, 4, 5, 6, 7$. The sum $a_2 + a_3 + a_4 + a_5 + a_6 + a_7$ is: