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. (a) 200
B. (b) 300
C. (c) 500
D. (d) 350

Answer

To solve the problem, we need to determine the number of different teams consisting of 2 girls and 3 boys that can be formed from a class of 5 girls and 7 boys, given that two specific boys (A and B) refuse to be in the same team.

### Step-by-Step Solution:

1. **Calculate Total Teams Without Restrictions:**
- We first calculate the total number of ways to choose 2 girls from 5 and 3 boys from 7 without any restrictions.
- The number of ways to choose 2 girls from 5 is given by the combination formula \( \binom{n}{r} \):
\[ \text{Number of ways to choose 2 girls} = \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
- The number of ways to choose 3 boys from 7 is:
\[ \text{Number of ways to choose 3 boys} = \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]
- Therefore, the total number of teams without restrictions is:
\[ \text{Total teams} = \binom{5}{2} \times \binom{7}{3} = 10 \times 35 = 350 \]

2. **Calculate Teams with Boys A and B Together:**
- Next, we calculate the number of teams where both boys A and B are included in the team.
- If A and B are both in the team, we need to choose 1 more boy from the remaining 5 boys (since A and B are already chosen):
\[ \text{Number of ways to choose 1 boy} = \binom{5}{1} = 5 \]
- The number of ways to choose 2 girls remains the same:
\[ \text{Number of ways to choose 2 girls} = \binom{5}{2} = 10 \]
- Therefore, the total number of teams with A and B together is:
\[ \text{Teams with A and B} = \binom{5}{2} \times \binom{5}{1} = 10 \times 5 = 50 \]

3. **Calculate Teams with Boys A and B Not Together:**
- To find the number of teams where A and B are not together, we subtract the number of teams with A and B together from the total number of teams:
\[ \text{Teams without A and B together} = \text{Total teams} - \text{Teams with A and B} = 350 - 50 = 300 \]

### Final Answer:
The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, with the restriction that boys A and B cannot be on the same team, is **300**.