Question
How many different committees of 5 students can be formed with 2 boys and 3 girls from 5 boys and 6 girls?
Answers
GPT-4o mini
To find the number of different committees of 5 students consisting of 2 boys and 3 girls from a group of 5 boys and 6 girls, we can use combinations.
1. **Choosing the Boys:**
We need to select 2 boys from the 5 available boys. The number of ways to choose 2 boys from 5 can be calculated using the combination formula \( \binom{n}{r} \), which is given by:
\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
For this situation:
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
\]
2. **Choosing the Girls:**
Next, we need to select 3 girls from the 6 available girls. We can also use the combination formula for this selection:
\[
\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20
\]
3. **Calculating the Total Committees:**
Now, to find the total number of different committees that can be formed, we multiply the number of ways to choose the boys and the number of ways to choose the girls:
\[
\text{Total Committees} = \binom{5}{2} \times \binom{6}{3} = 10 \times 20 = 200
\]
Thus, the total number of different committees of 5 students that can be formed with 2 boys and 3 girls is \( \boxed{200} \).
1. **Choosing the Boys:**
We need to select 2 boys from the 5 available boys. The number of ways to choose 2 boys from 5 can be calculated using the combination formula \( \binom{n}{r} \), which is given by:
\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
For this situation:
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
\]
2. **Choosing the Girls:**
Next, we need to select 3 girls from the 6 available girls. We can also use the combination formula for this selection:
\[
\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20
\]
3. **Calculating the Total Committees:**
Now, to find the total number of different committees that can be formed, we multiply the number of ways to choose the boys and the number of ways to choose the girls:
\[
\text{Total Committees} = \binom{5}{2} \times \binom{6}{3} = 10 \times 20 = 200
\]
Thus, the total number of different committees of 5 students that can be formed with 2 boys and 3 girls is \( \boxed{200} \).