$n$ balls each of weight $w$ when weighted in pairs the sum of the weights of all the possible pairs is $120$ when they are weighed in triplets the sum of the weights comes out to be $480$ for all possible triplets, then $n$ is

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. . . .

For $2 \le r \le n,\left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right) + 2\,\left( \begin{array}{l}\,\,n\\r - 1\end{array} \right)$ $ + \left( {\begin{array}{*{20}{c}}n\\{r - 2}\end{array}} \right)$ is equal to

$\left( {\begin{array}{*{20}{c}}n\\{n - r}\end{array}} \right)\, + \,\left( {\begin{array}{*{20}{c}}n\\{r + 1}\end{array}} \right)$, whenever $0 \le r \le n - 1$is equal to

A group of students comprises of $5$ boys and $n$ girls. If the number of ways, in which a team of $3$ students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is $1750$, then $n$ is equal to

If $^n{P_3}{ + ^n}{C_{n - 2}} = 14n$, then $n = $