$^n{P_r} \div ^n{C_r} = $

If $\frac{1}{8!} + \frac{1}{9!} = \frac{x}{10!},$ find $x.$

$A$ bag contains $n$ white and $n$ black balls. Pairs of balls are drawn at random without replacement successively,until the bag is empty. If the number of ways in which each pair consists of one white and one black ball is $14400$,then $n$ is equal to

If ${}^n P_r = 30240$ and ${}^n C_r = 252$,then the ordered pair $(n, r)$ is equal to

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

Let $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. Let $x$ be the number of $9$-digit numbers formed using the digits of the set $S$ such that only one digit is repeated and it is repeated exactly twice. Let $y$ be the number of $9$-digit numbers formed using the digits of the set $S$ such that only two digits are repeated and each of these is repeated exactly twice. Then,