$^{37}C_4 + \sum_{r=1}^{5} {^{(42-r)}C_r} = $

$A$ certain question paper contains three parts $A, B, C$ with four questions in part $A$,five questions in part $B$,and six questions in part $C$. $A$ student is required to answer seven questions,choosing at least two questions from each part. The total number of different ways a student can choose his seven questions for answering is:

Let $S = \{0, 1, 2, 3, \ldots, 100\}$. The number of ways of selecting $x, y \in S$ such that $x \neq y$ and $x + y = 100$ 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:

The total number of six-digit numbers formed using the digits $4, 5, 9$ only and divisible by $6$ is $.........$.

The number of ways in which $4$ different things can be distributed to $6$ persons so that no person gets all the things is