Find the number of ways in which two Americans, two British, One Chinese, One Dutch and one Egyptian can sit on a round table so that person of the same nationality are separated?

$^{20}C_1 + 3 ^{20}C_2 + 3 ^{20}C_3 + ^{20}C_4$ is equal to-

Number of integral solutions to the equation $x+y+z=21$, where $x \geq 1, y \geq 3, z \geq 4$, is equal to $..........$.

Let the number of elements in sets $A$ and $B$ be five and two respectively. Then the number of subsets of $A \times B$ each having at least $3$ and at most $6$ element is :

The number of onto functions $f$ from $\{1, 2, 3, …, 20\}$ only $\{1, 2, 3, …, 20\}$ such that $f(k)$ is a multiple of $3$, whenever $k$ is a multiple of $4$, is

$m$ men and $n$ women are to be seated in a row so that no two women sit together. If $m > n$, then the number of ways in which they can be seated is