If $^{n-1}C_r = (k^2 - 3) \cdot ^nC_{r+1}$,then an interval containing the values of $k$ is

In an election,a voter can vote for any number of candidates but not more than the number of candidates to be elected. There are $10$ candidates and $4$ are to be elected. If a voter must vote for at least one candidate,in how many ways can they vote?

$A$ student is to answer $10$ out of $13$ questions in an examination such that he must choose at least $4$ from the first $5$ questions. The number of choices available to him is

Consider three boxes,each containing $10$ balls labelled $1, 2, \dots, 10$. Suppose one ball is randomly drawn from each of the boxes. Denote by $n_i$ the label of the ball drawn from the $i^{th}$ box,$(i = 1, 2, 3)$. Then,the number of ways in which the balls can be chosen such that $n_1 < n_2 < n_3$ is:

The number of elements in the set $S = \{(r, k) : k \in Z \text{ and } ^{36}C_{r+1} = \frac{6(^{35}C_r)}{(k^2-3)}\}$ is:

The domain of definition of the function $f(x) = \log_{[x + \frac{1}{x}]} |x^2 - x - 6| + ^{16-x}C_{2x-1} + ^{20-3x}P_{2x-5}$ is,where $[x]$ denotes the greatest integer function.