In how many ways can $5$ boys and $5$ girls stand in a row so that no two girls may be together?

If a polygon of $n$ sides has $560$ diagonals,then $n=$

Let $l_1$ and $l_2$ be two lines intersecting at $P$. If $A_1, B_1, C_1$ are points on $l_1$,and $A_2, B_2, C_2, D_2, E_2$ are points on $l_2$,and if none of these points coincides with $P$,then the number of triangles formed by these eight points is:

Suppose $t_n$ is the number of triangles formed using the vertices of a regular polygon of $n$ sides. If $t_{n+1} = t_n + 28$,then $n =$

There are $10$ points in a plane of which no three points are collinear except $4$ points. The number of distinct triangles that can be formed by joining these points such that at least one of the vertices of every triangle formed is from the given $4$ collinear points is:

If $t_n$ denotes the number of triangles formed with $n$ points in a plane,no three of which are collinear,and if $t_{n+1}-t_n=36$,then $n$ is equal to