In how many ways can a committee of $6$ members be formed from $8$ men and $4$ women such that the committee contains at least $3$ women?

If $^nC_r = 84$,$^nC_{r-1} = 36$,and $^nC_{r+1} = 126$,then $n$ equals

If $\binom{n}{3} + \binom{n}{4} > \binom{n+1}{3}$,then:

If $\binom{n}{r-1} = 36$,$\binom{n}{r} = 84$,and $\binom{n}{r+1} = 126$,then $r = \dots$

Let $n$ and $k$ be positive integers such that $n \ge \frac{k(k + 1)}{2}$. The number of solutions $(x_1, x_2, ..., x_k)$ where $x_1 \ge 1, x_2 \ge 2, ..., x_k \ge k$ are all integers,satisfying $x_1 + x_2 + ... + x_k = n$,is

$A$ committee of $7$ members is to be formed from $9$ boys and $4$ girls. In how many ways can this be done if the committee must consist of exactly $3$ girls?