In how many ways can $4$ balls be picked from $6$ black and $4$ green colored balls such that at least one black ball is selected?

If $\binom{2n}{3} : \binom{n}{2} = 44 : 3$ and $\binom{n}{r} = 15$,then $r = \dots$

If at the end of a certain meeting,everyone had shaken hands with everyone else,and it was found that $45$ handshakes were exchanged,then the number of members present at the meeting is:

$A$ student has to answer $10$ questions,choosing at least $4$ from each of the parts $A$ and $B$. If there are $6$ questions in part $A$ and $7$ in part $B$,then the number of ways the student can choose $10$ questions is:

If ${}^nC_{r-1}=36$,${}^nC_r=84$,and ${}^nC_{r+1}=126$,then the value of $nr^2$ is

$A$ bag contains $5$ red marbles,$4$ black marbles,and $3$ white marbles. The number of ways in which $4$ marbles can be drawn so that at most $2$ of them are red is: