Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 bo

Q: Why can't we just pick the boys first, then the girls?

You absolutely can! The order of selection doesn't change the final count because the choices are independent. Multiplication is commutative, so whether you calculate (Girls $\times$ Boys) or (Boys $\times$ Girls), the result remains the same.

Q: Why do we subtract the forbidden cases instead of adding the valid ones?

Adding valid cases (A only, B only, neither) is much harder because you have to calculate three separate scenarios. Subtracting the forbidden case (A and B together) is just one calculation. It's the path of least resistance!

Q: What if the team size was different?

The logic remains the same. The 'Total - Forbidden' strategy is a universal tool in combinatorics. Just identify the 'forbidden' configuration and subtract it from the total. The core principle is to simplify the problem by removing the 'noise' of the constraint.

Question

Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is:

Options:

(A) 200

(B) 300

(C) 500

(D) 350

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with selecting a team of 5 members from a pool of 5 girls and 7 boys. The team must consist of exactly 2 girls and 3 boys.

However, there is a specific constraint: two boys, $A$ and $B$, refuse to work together. We must calculate the number of valid team combinations that satisfy these conditions.

### The Power of Complementary Counting

When faced with a constraint like "A and B cannot be together," calculating every valid scenario directly is tedious and prone to error. Instead, we use **Complementary Counting**.

The strategy is to calculate the total number of ways to form the team without any restrictions, and then subtract the "forbidden" scenarios where both $A$ and $B$ are present.

### Phase 1: The Unconstrained Total

First, we calculate the total number of ways to choose 2 girls from 5 and 3 boys from 7, ignoring the conflict between $A$ and $B$.

The number of ways to choose 2 girls from 5 is:

$$^{5}C_{2} = \frac{5 	imes 4}{2 	imes 1} = 10$$

The number of ways to choose 3 boys from 7 is:

$$^{7}C_{3} = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$$

Since these selections are independent, the total number of unconstrained teams is:

$$10 	imes 35 = 350$$

### Phase 2: The Forbidden Scenario

Next, we isolate the forbidden teams where both $A$ and $B$ are included. If $A$ and $B$ are already on the team, they occupy 2 of the 3 available boy slots.

We only need to select 1 more boy from the remaining 5 boys ($7 - 2 = 5$). The number of ways to choose this final boy is:

$$^{5}C_{1} = 5$$

For each of these 5 boy combinations, we still have the original 10 ways to choose the girls. Therefore, the total number of forbidden teams is:

$$10 	imes 5 = 50$$

### Phase 3: The Final Resolution

Finally, we subtract the forbidden scenarios from the total universe of possible teams to find the valid count.

$$350 - 50 = 300$$

By using the principle of complementary counting, we have determined that there are **300** valid ways to form the team. Remember, in combinatorics, the most complex problems often yield to the simplest, most logical strategies.

Animated Solution for Mathematics - Permutations and Combinations: Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is:

Select Answer:

Visualized Solution

The Selection Pool

Selecting the Girls

Calculating Girl Combinations

Selecting Boys (Ignoring Constraints)

Calculating Boy Combinations

Total Unconstrained Teams

The Forbidden Scenario

Forcing A and B into the Team

Completing the Forbidden Boy Group

Total Forbidden Teams

The Final Calculation

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Power of Complementary Counting

Phase 1

Phase 2

Phase 3

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