Question

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

• A. $P\to 4;Q\to 6;R\to 2;S\to 1$
• B. $P\to 1;Q\to 4;R\to 2;S\to 3$
• C. $P\to 4;Q\to 6;R\to 5;S\to 2$
• D. $P\to 4;Q\to 2;R\to 3;S\to 1$

Answer 1

The correct option is C $P\to 4;Q\to 6;R\to 5;S\to 2$

(1) ${\alpha }_1=(\frac{6}{3})(\frac{5}{2})=200$
So $P\to 4$
(2) ${\alpha }_2=(\frac{6}{1})(\frac{5}{1})+(\frac{6}{2})(\frac{5}{2})+(\frac{6}{3})(\frac{5}{3})+(\frac{6}{4})(\frac{5}{4})+(\frac{6}{5})(\frac{5}{5})$
$=30+150+200+75+6$
$=461$
So $Q\to 6$
(3) ${\alpha }_3=(\frac{5}{2})(\frac{6}{3})+(\frac{5}{3})(\frac{6}{2})+(\frac{5}{4})(\frac{6}{1})+(\frac{5}{5})(\frac{6}{0})$
$=(\frac{11}{5})-(\frac{5}{0})(\frac{6}{5})-(\frac{5}{1})(\frac{6}{4})$
$=381$
So $R\to 5$
(4) ${\alpha }_2=(\frac{5}{2})(\frac{6}{2})-(\frac{4}{1})(\frac{5}{1})+(\frac{5}{3})(\frac{6}{1})-(\frac{4}{2})(\frac{5}{0})+(\frac{5}{4})=189$
So $S\to 2$.