The number of solutions of the equation $x+y+z=10$,where $x, y$,and $z$ are positive integers,is:

$A$ student is allowed to select at most $n$ books from a collection of $(2n+1)$ books. If the total number of ways in which one can select at least one book is $255$,then the value of $n$ is

$A$ group consists of $4$ girls and $7$ boys. In how many ways can a team of $5$ members be selected if the team has at least one boy and one girl?

In how many ways can Rs. $16$ be divided among $4$ persons if none of them gets less than Rs. $3$?

$A$ group of $9$ students,$s_1, s_2, \ldots, s_9$,is to be divided into three teams $X, Y$,and $Z$ of sizes $2, 3$,and $4$,respectively. Suppose that $s_1$ cannot be selected for team $X$,and $s_2$ cannot be selected for team $Y$. The number of ways to form such teams is:

If the number of $5$-element subsets of the set $A = \{a_1, a_2, \dots, a_{20}\}$ of $20$ distinct elements is $k$ times the number of $5$-element subsets containing $a_4$,then $k$ is