If $^{n}C_{9} = ^{n}C_{8}$,find $^{n}C_{17}$.

Compute $\frac{8!}{6! \times 2!}$

If $\binom{189}{35} + \binom{189}{x} = \binom{190}{x}$,then $x = \dots$

If $2 \times {}^nC_5 = 9 \times {}^{n-2}C_5$,then the value of $n$ is:

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) = $

Determine $n$ if $^{2n}C_{3} : ^{n}C_{3} = 12 : 1$.