If $^{n + 1}C_3 = 2 \cdot ^nC_2$,then $n =$

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

$^{2n}C_4 : ^nC_3 = 99 : 4 \Rightarrow n = $

$A$ lady gives a dinner party for $6$ guests. The number of ways in which they may be selected from among $10$ friends,if two of the friends will not attend the party together,is:

Two players $A$ and $B$ play a series of games of badminton. The player,who wins $5$ games first,wins the series. Assuming that no game ends in a draw,the number of ways,in which player $A$ wins the series is . . . . . . .

If $^{20}C_{n+2} = ^nC_{16}$,then the value of $n$ is