The number of ways, in which $5$ girls and $7$ boys can be seated at a round table so that no two girls sit together, is
$126(5 !)^2$
$7(360)^2$
$720$
$7(720)^2$
Value of $r$ for which $^{15}{C_{r + 3}} = {\,^{15}}{C_{2r - 6}}$ is
If $\alpha { = ^m}{C_2}$, then $^\alpha {C_2}$is equal to
Team $'A'$ consists of $7$ boys and $n$ girls and Team $'B'$ has $4$ boys and $6$ girls. If a total of $52$ single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then $n$ is equal to
A group of $9$ students, $s 1, s 2, \ldots, s 9$, is to be divided to form three teams $X, Y$ and, $Z$ of sizes $2,3$ , and $4$, respectively. Suppose that $s_1$ cannot be selected for the team $X$, and $s_2$ cannot be selected for the team $Y$. Then the number of ways to form such teams, is. . . .