In a circus,there are $10$ cages for accommodating $10$ animals. Out of these,$4$ cages are so small that $5$ out of the $10$ animals cannot enter them. In how many ways is it possible to accommodate the $10$ animals in these $10$ cages?

Let $n$ be the number of ways in which $5$ boys and $5$ girls can stand in a queue such that all the girls stand consecutively. Let $m$ be the number of ways in which $5$ boys and $5$ girls can stand in a queue such that exactly four girls stand consecutively. Then the value of $\frac{m}{n}$ is

Let $a_1, a_2, \ldots, a_n$ be $n$ non-zero real numbers,of which $p$ are positive and the remaining are negative. The number of ordered pairs $(j, k)$ with $j < k$ for which $a_j a_k$ is positive is $55$. Similarly,the number of ordered pairs $(j, k)$ with $j < k$ for which $a_j a_k$ is negative is $50$. Then,the value of $p^2 + (n-p)^2$ is

The total number of two-digit numbers $n$ such that $3^{n} + 7^{n}$ is a multiple of $10$ is ..... .

If $^nC_r = ^nC_{r-1}$ and $^nP_r = ^nP_{r+1}$,then the value of $n$ is

$A$ building has a ground floor and $10$ more floors. Nine persons enter a lift at the ground floor. The lift goes up to the $10$th floor. The number of ways in which any $4$ persons exit at a floor and the remaining $5$ persons exit at a different floor, if the lift does not stop at the first and the second floors, is equal to: