In the product $(1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) \dots (1 + x + x^2 + \dots + x^{100})$,when written in ascending powers of $x$,the highest exponent of $x$ is . . . . . . .

Let $m$ (respectively,$n$) be the number of $5$-digit integers obtained by using the digits $1, 2, 3, 4, 5$ with repetitions (respectively,without repetitions) such that the sum of any two adjacent digits is odd. Then $\frac{m}{n}$ is equal to

The smallest positive integer $n$,for which $n! < {\left( {\frac{{n + 1}}{2}} \right)^n}$ holds 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:

$5$ boys and $6$ girls are arranged in all possible ways. Let $X$ denote the number of linear arrangements in which no two boys sit together and $Y$ denote the number of linear arrangements in which no two girls sit together. If $Z$ denotes the number of ways of arranging all of them around a circular table such that no two boys sit together,then $X: Y: Z=$

If a number is chosen at random from the four-digit numbers formed by using the digits $0, 1, 2, 3, 4, 6$ without repetition,then the probability that it is divisible by $4$ is