Consider the following statements:
$I$. The number of positive integral solutions of $x_1+x_2+x_3+x_4=10$ is $286$.
$II$. If $25! = 10^n \times k, (k \in N)$,then $n=6$.
Which one of the following options is true?

If $f(n) = n! (31-n)!$,where $n \in \{0, 1, 2, \ldots, 31\}$,then the minimum value of $f(n)$ is

Five balls of different colours are to be placed in three boxes of different sizes. The number of ways in which we can place the balls in the boxes so that no box remains empty is

$6$ different letters of an alphabet are given. Words with $4$ letters are formed from these given letters. The number of words which have at least one letter repeated and no two same letters are together is:

In a high school,a committee has to be formed from a group of $6$ boys $M_1, M_2, M_3, M_4, M_5, M_6$ and $5$ girls $G_1, G_2, G_3, G_4, G_5$.
$(i)$ Let $\alpha_1$ be the total number of ways in which the committee can be formed such that the committee has $5$ members,having exactly $3$ boys and $2$ girls.
$(ii)$ Let $\alpha_2$ be the total number of ways in which the committee can be formed such that the committee has at least $2$ members,and having an equal number of boys and girls.
$(iii)$ Let $\alpha_3$ be the total number of ways in which the committee can be formed such that the committee has $5$ members,at least $2$ of them being girls.
$(iv)$ Let $\alpha_4$ be the total number of ways in which the committee can be formed such that the committee has $4$ members,having at least $2$ girls and such that both $M_1$ and $G_1$ are $NOT$ in the committee together.

$LIST-I$	$LIST-II$
$P$. The value of $\alpha_1$ is	$1. 136$
$Q$. The value of $\alpha_2$ is	$2. 189$
$R$. The value of $\alpha_3$ is	$3. 192$
$S$. The value of $\alpha_4$ is	$4. 200$
	$5. 381$
	$6. 461$

The correct option is:

The sum of all the four-digit numbers that can be formed using all the digits $2, 1, 2, 3$ is equal to $.......$.