In how many ways can $5$ boys and $5$ girls be arranged in a row such that no two girls are together?

The number of diagonals of a polygon is $35$. If $A$ and $B$ are two distinct vertices of this polygon,then the number of all those triangles formed by joining three vertices of the polygon having $AB$ as one of its sides is:

Let $t_n$ denote the number of integral-sided triangles with distinct sides chosen from $\{1, 2, 3, \ldots, n\}$. Then,$t_{20} - t_{19}$ equals

In the given figure,$6$ '$A$'s should be written in such a manner that every row contains at least one '$A$'. The number of ways this is possible is

Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball,the second row consists of two balls and so on. If $99$ more identical balls are added to the total number of balls used in forming the equilateral triangle,then all these balls can be arranged in a square whose each side contains exactly $2$ balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is

$A$ polygon has $54$ diagonals. The number of sides of this polygon is