1. Compute $\alpha_1$

We need a 5-member committee with exactly 3 boys and 2 girls.

From 6 boys, choose 3: $\binom{6}{3}=20$

From 5 girls, choose 2: $\binom{5}{2}=10$

Hence, $\alpha_1=\binom{6}{3}\binom{5}{2}=20\cdot 10=200$

So, in List-II: $P \to 4$

2. Compute $\alpha_2$

Committee has at least 2 members and equal number of boys and girls.

If the number of boys equals the number of girls, let each be $r$. Then total members $=2r$.

Since there are 5 girls, $r$ can be: $r=1,2,3,4,5$

Now count for each case:

• For $r=1$: $\binom{6}{1}\binom{5}{1}=6\cdot 5=30$

• For $r=2$: $\binom{6}{2}\binom{5}{2}=15\cdot 10=150$

• For $r=3$: $\binom{6}{3}\binom{5}{3}=20\cdot 10=200$

• For $r=4$: $\binom{6}{4}\binom{5}{4}=15\cdot 5=75$

• For $r=5$: $\binom{6}{5}\binom{5}{5}=6\cdot 1=6$

Therefore, $\alpha_2=30+150+200+75+6=461$

So, $Q \to 6$

3. Compute $\alpha_3$

We need a 5-member committee with at least 2 girls.

Possible girl counts are 2, 3, 4, 5.

Case 1: 2 girls, 3 boys

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

Case 2: 3 girls, 2 boys

$\binom{5}{3}\binom{6}{2}=10\cdot 15=150$

Case 3: 4 girls, 1 boy

$\binom{5}{4}\binom{6}{1}=5\cdot 6=30$

Case 4: 5 girls, 0 boys

$\binom{5}{5}\binom{6}{0}=1\cdot 1=1$

Thus, $\alpha_3=200+150+30+1=381$

So, $R \to 5$

4. Compute $\alpha_4$

We need a 4-member committee with at least 2 girls, and both $M_1$ and $G_1$ are not together.

First count all 4-member committees with at least 2 girls. Then subtract those containing both $M_1$ and $G_1$.

Total committees with at least 2 girls

Possible compositions:

• 2 girls, 2 boys
• 3 girls, 1 boy
• 4 girls, 0 boys

So, $\binom{5}{2}\binom{6}{2}+\binom{5}{3}\binom{6}{1}+\binom{5}{4}\binom{6}{0}$ $=10\cdot 15+10\cdot 6+5\cdot 1$ $=150+60+5=215$

Committees containing both $M_1$ and $G_1$

We already have 2 members fixed: $M_1, G_1$. Need 2 more members, with overall at least 2 girls.

Now count valid ways:

Subcase A: total 2 girls, 2 boys

Since $G_1$ is already one girl and $M_1$ one boy, choose:

• 1 more girl from remaining 4 girls
• 1 more boy from remaining 5 boys

Count: $\binom{4}{1}\binom{5}{1}=4\cdot 5=20$

Subcase B: total 3 girls, 1 boy

Need 2 more girls from remaining 4 girls: $\binom{4}{2}=6$

Subcase C: total 4 girls, 0 boys

Impossible because $M_1$ is already included.

Hence committees containing both $M_1$ and $G_1$: $20+6=26$

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

So, $S \to 2$

5. Match with List-II

We obtained:

• $\alpha_1=200 \Rightarrow P \to 4$
• $\alpha_2=461 \Rightarrow Q \to 6$
• $\alpha_3=381 \Rightarrow R \to 5$
• $\alpha_4=189 \Rightarrow S \to 2$

This matches: $\boxed{\text{Option C}}$

6. Comparison with stored correct answer

Stored correct answer = C.

Our derived answer is also C, so they agree.