How many $5$-digit numbers can be formed using the digits $0, 1, 2, 3, 4,$ and $5$ without repetition such that the number is divisible by $3$?

$A$ test containing $3$ objective-type questions is conducted in a class. Each question has $4$ options and only one option is the correct answer. No two students of the class have answered identically,and no student has written all correct answers. If every student has attempted all the questions,then the maximum possible number of students who have written the test is:

The number of positive integers less than $10000$ which contain the digit $5$ at least once 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 the number of circular permutations of $9$ distinct things taken $5$ at a time is $n_1$ and the number of linear permutations of $8$ distinct things taken $4$ at a time is $n_2$,then $\frac{n_1}{n_2}=$

In the prime factorization of $37! = 2^{\alpha_2} \cdot 3^{\alpha_3} \cdot 5^{\alpha_5} \cdots 37^{\alpha_{37}}$,the ratio $\alpha_3 : \alpha_5$ is: