Two packs of $52$ cards are shuffled together. The number of ways in which a man can be dealt $26$ cards so that he does not get two cards of the same suit and same denomination is

Match the items of List-$I$ to the items of List-$II$:

	List-$I$		List-$II$
$(A)$	The number of ways of not selecting $(n-r)$ things from $n$ different things	$(I)$	$1+n+{ }^n C_2+\ldots+{ }^n C_r$
$(B)$	$(n-r+1) \cdot{ }^n C_{r-1}$	$(II)$	$(r+1) \cdot{ }^n C_{r+1}$
$(C)$	The number of ways of selecting at least $(n-r)$ things from $n$ different things	$(III)$	$r\left({ }^n C_r\right)$
$(D)$	$(n-r)\left({ }^{n-1} C_{r-1}+{ }^{n-1} C_r\right)$	$(IV)$	$2^n-1-n-{ }^n C_2-\ldots-{ }^n C_r$
		$(V)$	${ }^n C_{n-r}$

The correct match is:

If $P(n, r) = 1680$ and $C(n, r) = 70$,then $69n + r! = \dots$.

$A$ 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:

The number of positive integral solutions of the equation $xyz = 3000$ is

Let $n \geq 3$ be an integer. For a permutation $\sigma = (a_1, a_2, \ldots, a_n)$ of $(1, 2, \ldots, n)$,we define $f_\sigma(x) = a_n x^{n-1} + a_{n-1} x^{n-2} + \ldots + a_2 x + a_1$. Let $S_\sigma$ be the sum of the roots of $f_\sigma(x) = 0$ and let $S$ denote the sum over all permutations $\sigma$ of $(1, 2, \ldots, n)$ of the values $S_\sigma$. Then,