In a high school, a committee has to be formed from a group of 6 boys M1, M2, M3, M4, M5, M6 and 5 g

Q: Wait, why do we subtract 1 in the calculation of $a_2$?

That is a brilliant observation! Vandermonde's identity, $\sum_{k=0}^{n} ^{r}C_{k} \times ^{m}C_{n-k} = ^{r+m}C_{n}$, sums from $k=0$ to $n$. In our case, $k=0$ corresponds to selecting 0 boys and 0 girls. This results in an empty committee. Since the problem explicitly states the committee must have at least 2 members, we must remove that $k=0$ case from our total.

Q: Is complementary counting always better for 'at least' problems?

Not always, but it is a powerful tool. When you see 'at least', check the number of cases. If calculating the 'at least' cases involves 4 or 5 scenarios, but the 'complementary' cases involve only 1 or 2, then complementary counting is definitely the faster, safer route. It minimizes the chance of calculation errors.

Q: How do I handle the 'not together' constraint in $a_4$ without getting confused?

The 'not together' constraint is a classic trap. The most robust method is: Total Valid Committees (ignoring the constraint) minus Violating Committees (where they are together). When calculating the violating cases, treat M1 and G1 as 'already selected'. This reduces the number of spots to fill and reduces the pool of available students. It turns a complex constraint into a simple selection problem.

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 $a_1$ 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 $a_2$ 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 $a_3$ 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 $a_4$ be the total number of ways in which the committee can be formed such that the committee has $4$ members, having atleast $2$ girls and such that both $M_1$ and $G_1$ are NOT in the committee together. Match the values in List-I to the numbers in List-II.

List-I:

P. The value of $a_1$ is

Q. The value of $a_2$ is

R. The value of $a_3$ is

S. The value of $a_4$ is

List-II:

1. 136

2. 189

3. 192

4. 200

5. 381

6. 461

Sigma Percentile · Accepted Answer

### The Art of Selection: Mastering Combinatorics

Welcome, future engineers! Today, we are not just solving a combinatorics problem; we are learning to see the hidden structures in a crowd. Combinatorics is the mathematics of choice, and in the JEE, it is often less about brute force and more about finding the elegant path.

Let us break down this committee problem step by step.

### Phase 1: The Foundation ($a_1$)

We begin with a simple task: forming a committee of 5 members with exactly 3 boys and 2 girls. We have 6 boys and 5 girls. This is a direct application of the Fundamental Principle of Counting.

We treat the selection of boys and girls as independent events. We choose 3 boys from 6, which is $^{6}C_{3}$, and 2 girls from 5, which is $^{5}C_{2}$.

Multiplying these gives us the total ways:

$$a_1 = ^{6}C_{3} 	imes ^{5}C_{2} = 20 	imes 10 = 200$$

### Phase 2: The Elegance of Vandermonde ($a_2$)

Now, things get interesting. We need an equal number of boys and girls, with at least 2 members total. If we have $k$ boys and $k$ girls, the total members are $2k$. Since we have 5 girls, $k$ can range from 1 to 5.

The total ways would be:

$$\sum_{k=1}^{5} (^{6}C_{k} 	imes ^{5}C_{k})$$

Calculating this sum directly is a recipe for a headache. Instead, let us invoke Vandermonde's Identity:

$$\sum_{k=0}^{n} ^{r}C_{k} 	imes ^{m}C_{n-k} = ^{r+m}C_{n}$$

By rewriting $^{5}C_{k}$ as $^{5}C_{5-k}$, our sum becomes $\sum_{k=0}^{5} (^{6}C_{k} 	imes ^{5}C_{5-k})$. This is exactly $^{11}C_{5}$.

But wait! Our sum starts at $k=1$, not $k=0$. So, we must subtract the $k=0$ case (the empty committee).

$$a_2 = ^{11}C_{5} - (^{6}C_{0} 	imes ^{5}C_{0}) = 462 - 1 = 461$$

### Phase 3: The Power of Complementary Counting ($a_3$)

For $a_3$, we need a 5-member committee with at least 2 girls. The phrase 'at least' is a siren song in JEE problems—it often lures students into calculating too many cases. Let us use complementary counting.

We take the total possible 5-member committees and subtract the 'unwanted' cases: those with 0 girls or 1 girl.

Total ways = $^{11}C_{5} = 462$.

Ways with 0 girls = $^{6}C_{5} 	imes ^{5}C_{0} = 6 	imes 1 = 6$.

Ways with 1 girl = $^{6}C_{4} 	imes ^{5}C_{1} = 15 	imes 5 = 75$.

Subtracting these from the total, we get:

$$a_3 = 462 - (6 + 75) = 381$$

### Phase 4: The Boss Level ($a_4$)

Finally, we tackle $a_4$: a 4-member committee, at least 2 girls, with the constraint that $M_1$ and $G_1$ cannot be together. We use the same 'Total minus Violating' strategy.

First, find the total 4-member committees with at least 2 girls (ignoring the constraint):

- 2 Girls, 2 Boys: $^{5}C_{2} 	imes ^{6}C_{2} = 10 	imes 15 = 150$
- 3 Girls, 1 Boy: $^{5}C_{3} 	imes ^{6}C_{1} = 10 	imes 6 = 60$
- 4 Girls, 0 Boys: $^{5}C_{4} 	imes ^{6}C_{0} = 5 	imes 1 = 5$

Total = $150 + 60 + 5 = 215$.

Now, subtract the violating cases where $M_1$ and $G_1$ are both present. We have already filled 2 spots ($M_1, G_1$). We need 2 more members from the remaining 5 boys and 4 girls, ensuring at least 1 more girl (to satisfy the 'at least 2 girls' rule):

- 1 more Girl, 1 more Boy: $^{4}C_{1} 	imes ^{5}C_{1} = 4 	imes 5 = 20$
- 2 more Girls, 0 more Boys: $^{4}C_{2} 	imes ^{5}C_{0} = 6 	imes 1 = 6$

Total violations = $20 + 6 = 26$.

Finally, we arrive at the result:

$$a_4 = 215 - 26 = 189$$

You have navigated the constraints, applied the identities, and conquered the problem. Keep this mindset, and no JEE problem will ever intimidate you again!

List-I

List-II

Select Matching Pairs:

Visualized Solution

Initial Setup

Condition for $a_{1}$

Computing $a_{1}$

Condition for $a_{2}$

Simplifying $a_{2}$

Computing $a_{2}$

Condition for $a_{3}$

Computing Unwanted Cases

Computing $a_{3}$

Condition for $a_{4}$

Total without Restriction

The Violating Cases

Computing Violations

Computing $a_{4}$

Final Matching

The Sigma Insight: Combinations and Selection

The Art of Selection

Phase 1

Phase 2

Phase 3

Phase 4

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