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 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 |

$6 \text{ boys } M_1,\dots,M_6;\qquad 5 \text{ girls } G_1,\dots,G_5$

$\alpha_1:\ 5 \text{ members, exactly } 3 \text{ boys and } 2 \text{ girls}$

$\alpha_2:\ \text{at least } 2 \text{ members, equal boys and girls}$

$\alpha_3:\ 5 \text{ members, at least } 2 \text{ girls}$

$\alpha_4:\ 4 \text{ members, } \geq 2 \text{ girls},\ M_1 \text{ and } G_1 \text{ not together}$

$\text{To find: } \alpha_1,\alpha_2,\alpha_3,\alpha_4 \text{, then match List-I with List-II}$

$\alpha_1=\binom{6}{3}\binom{5}{2}$

$=20\times 10=200\ \Rightarrow\ \alpha_1\to(4)$

$\alpha_2:\ k \text{ boys} + k \text{ girls},\ k=1,2,3,4,5$

$\alpha_2=\sum_{k=1}^{5}\binom{6}{k}\binom{5}{k}$

$\alpha_2=\binom61\binom51+\binom62\binom52+\binom63\binom53+\binom64\binom54+\binom65\binom55$

$=30+150+200+75+6=461$

$\Rightarrow\ \alpha_2\to(6)$

$\text{total 5-member committees } = \binom{11}{5}=462$

$\text{subtract those with } <2 \text{ girls}$

$\binom60\binom65=6,\qquad \binom51\binom64=5\times 15=75$

$\alpha_3=462-6-75=381\ \Rightarrow\ \alpha_3\to(5)$

$\alpha_4:\ 4 \text{ members, } \geq 2 \text{ girls, not both } M_1,G_1$

$\text{first count } \geq 2 \text{ girls}: \binom52\binom62+\binom53\binom61+\binom54\binom60$

$=150+60+5=215$

$\text{both } M_1,G_1:\ \binom92-\binom52=36-10=26$

$\alpha_4=215-26=189\ \Rightarrow\ \alpha_4\to(2)$

$\alpha_1=200\to4,\qquad \alpha_2=461\to6$

$\alpha_3=381\to5,\qquad \alpha_4=189\to2$

$\text{Answer: } (3)\ P\to 4;\ Q\to 6;\ R\to 5;\ S\to 2$