If $\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}$,find $x$.

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 $x$ and $y$ represent the number of arrangements of the letters of the word $ATRAPATRAM$ such that $(i)$ all $A$'s are together and $(ii)$ no two $A$'s are together respectively,then $x+y$ is equal to:

The number of ways of selecting $8$ letters from $24$ letters,of which $8$ are $a$,$8$ are $b$,and the remaining $8$ are distinct,is given by:

The number of numbers between $2000$ and $5000$ that can be formed with the digits $0, 1, 2, 3, 4$ (repetition of digits not allowed) and are multiples of $3$ is:

The number of positive integral solutions of the equation $xyz = 90$ is equal to :-