If $^nC_{12} = ^nC_6$,then $^nC_2 = $

The number of ways in which $15$ identical gold coins can be distributed among $3$ persons such that each one gets at least $3$ gold coins is

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

$\binom{15}{8} + \binom{15}{9} - \binom{15}{6} - \binom{15}{7} = \dots$

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

$^{n-1}C_r = (k^2 - 8) ^nC_{r+1}$ if and only if: