Ways to Form Committee from Boys and Girls

To find the values of $\alpha_1, \alpha_2, \alpha_3,$ and $\alpha_4$, we apply combinatorial principles based on selecting members from $6$ boys ($M_1, M_2, M_3, M_4, M_5, M_6$) and $5$ girls ($G_1, G_2, G_3, G_4, G_5$).

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1. Calculation of $\alpha_1$ (P):

$\alpha_1$ is the number of ways to form a committee of $5$ members containing exactly $3$ boys and $2$ girls.

• Select $3$ boys out of $6$: $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$
• Select $2$ girls out of $5$: $\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$

$\alpha_1 = \binom{6}{3} \times \binom{5}{2} = 20 \times 10 = 200$

Thus, $\mathbf{P} \rightarrow 4$.

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2. Calculation of $\alpha_2$ (Q):

$\alpha_2$ is the number of ways to select a committee of at least $2$ members with equal numbers of boys and girls.

Let $k$ be the number of boys and girls selected, where $k \ge 1$ (so total members $2k \ge 2$). Since there are $5$ girls, $k \in \{1, 2, 3, 4, 5\}$.

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

• For $k = 1$: $\binom{6}{1} \binom{5}{1} = 6 \times 5 = 30$
• For $k = 2$: $\binom{6}{2} \binom{5}{2} = 15 \times 10 = 150$
• For $k = 3$: $\binom{6}{3} \binom{5}{3} = 20 \times 10 = 200$
• For $k = 4$: $\binom{6}{4} \binom{5}{4} = 15 \times 5 = 75$
• For $k = 5$: $\binom{6}{5} \binom{5}{5} = 6 \times 1 = 6$

Summing these values: $\alpha_2 = 30 + 150 + 200 + 75 + 6 = 461$

Thus, $\mathbf{Q} \rightarrow 6$.

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3. Calculation of $\alpha_3$ (R):

$\alpha_3$ is the number of ways to select a $5$-member committee having a minimum of $2$ girls.

The number of girls $g$ can be $2, 3, 4,$ or $5$, and the remaining members $(5 - g)$ are boys:

• $g = 2$ girls and $3$ boys: $\binom{5}{2} \binom{6}{3} = 10 \times 20 = 200$
• $g = 3$ girls and $2$ boys: $\binom{5}{3} \binom{6}{2} = 10 \times 15 = 150$
• $g = 4$ girls and $1$ boy: $\binom{5}{4} \binom{6}{1} = 5 \times 6 = 30$
• $g = 5$ girls and $0$ boys: $\binom{5}{5} \binom{6}{0} = 1 \times 1 = 1$

Summing these values: $\alpha_3 = 200 + 150 + 30 + 1 = 381$

Thus, $\mathbf{R} \rightarrow 5$.

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4. Calculation of $\alpha_4$ (S):

$\alpha_4$ is the number of ways to select a $4$-member committee containing at least $2$ girls such that $M_1$ and $G_1$ are NOT selected together.

Total ways without restriction:

• $2$ girls and $2$ boys: $\binom{5}{2} \binom{6}{2} = 10 \times 15 = 150$
• $3$ girls and $1$ boy: $\binom{5}{3} \binom{6}{1} = 10 \times 6 = 60$
• $4$ girls and $0$ boys: $\binom{5}{4} \binom{6}{0} = 5 \times 1 = 5$

$\text{Total ways} = 150 + 60 + 5 = 215$

Ways in which BOTH $M_1$ and $G_1$ ARE selected: Since $M_1$ and $G_1$ are both included, we need to pick $2$ more members from the remaining $5$ boys and $4$ girls such that the committee has at least $2$ girls in total (meaning at least $1$ more girl from the remaining $4$ girls):

• $1$ girl and $1$ boy from remaining: $\binom{4}{1} \binom{5}{1} = 4 \times 5 = 20$
• $2$ girls and $0$ boys from remaining: $\binom{4}{2} \binom{5}{0} = 6 \times 1 = 6$

$\text{Forbidden ways} = 20 + 6 = 26$

Therefore: $\alpha_4 = 215 - 26 = 189$

Thus, $\mathbf{S} \rightarrow 2$.

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Conclusion:

The correct matching is: $\mathbf{P} \rightarrow 4;\ \mathbf{Q} \rightarrow 6;\ \mathbf{R} \rightarrow 5;\ \mathbf{S} \rightarrow 2$

Hence, the correct option is C.