In how many ways can a committee of $6$ members be formed out of $10$ members,such that it always includes a specified member?

There are $7$ identical white balls and $3$ identical black balls. The number of distinguishable arrangements in a row of all the balls,so that no two black balls are adjacent is

$A$ committee of $11$ members is to be formed from $8$ males and $5$ females. If $m$ is the number of ways the committee is formed with at least $6$ males and $n$ is the number of ways the committee is formed with at least $3$ females,then:

Let $S_r = \{(x, y, z) : x + y + z = 11, x \geq r, y \geq r, z \geq r, x, y, z, r \in \mathbb{Z}\}$ and $n(S_r)$ represents the number of elements in $S_r$. Then $n(S_2) + n(S_3) + n(S_4) = $

Sangeeta plans a dinner party for $6$ guests. Out of $10$ friends,in how many ways can she choose the guests if two specific friends cannot attend the party together?

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