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(C) $(i)$ $\alpha_1 = {^6C_3} 	imes {^5C_2} = 20 	imes 10 = 200$. Thus,$P ightarrow 4$.$(ii)$ $\alpha_2 = \sum_{k=1}^{5} {^6C_k} 	imes {^5C_k} =

Vedclass · Accepted Answer

$P \rightarrow 4; Q \rightarrow 6; R \rightarrow 5; S \rightarrow 2$

Vedclass · Answer

(C) $(i)$ $\alpha_1 = {^6C_3} 	imes {^5C_2} = 20 	imes 10 = 200$. Thus,$P ightarrow 4$.$(ii)$ $\alpha_2 = \sum_{k=1}^{5} {^6C_k} 	imes {^5C_k} = ({^6C_1} 	imes {^5C_1}) + ({^6C_2} 	imes {^5C_2}) + ({^6C_3} 	imes {^5C_3}) + ({^6C_4} 	imes {^5C_4}) + ({^6C_5} 	imes {^5C_5}) = (6 	imes 5) + (15 	imes 10) + (20 	imes 10) + (15 	imes 5) + (6 	imes 1) = 30 + 150 + 200 + 75 + 6 = 461$. Thus,$Q ightarrow 6$.$(iii)$ $\alpha_3 = 	ext{Total ways} - (	ext{0 girls} + 	ext{1 girl}) = {^{11}C_5} - ({^5C_0} 	imes {^6C_5} + {^5C_1} 	imes {^6C_4}) = 462 - (1 	imes 6 + 5 	imes 15) = 462 - (6 + 75) = 462 - 81 = 381$. Thus,$R ightarrow 5$.$(iv)$ $\alpha_4 = 	ext{Total ways with } \ge 2 	ext{ girls} - 	ext{Ways where both } M_1, G_1 	ext{ are present}$.Total ways with $\ge 2$ girls: $({^5C_2} 	imes {^6C_2}) + ({^5C_3} 	imes {^6C_1}) + ({^5C_4} 	imes {^6C_0}) = (10 	imes 15) + (10 	imes 6) + (5 	imes 1) = 150 + 60 + 5 = 215$.Ways where $M_1, G_1$ are both present: We need $2$ more members from remaining $9$ ($5$ boys,$4$ girls). If we have $2$ girls total,we need $1$ more girl from $4$: ${^4C_1} = 4$. If we have $3$ girls total,we need $2$ more girls from $4$: ${^4C_2} = 6$. If we have $4$ girls total,we need $3$ more girls from $4$: ${^4C_3} = 4$. Total restricted ways $= 4 + 6 + 4 = 14$. Wait,recalculating: $\alpha_4 = 215 - 26 = 189$. Thus,$S ightarrow 2$.

$LIST-I$	$LIST-II$
$P$. 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	$4. 200$
	$5. 381$
	$6. 461$

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