There are 4 boys and 5 girls and you need to form a committee of 2 men and 3 women. In how many ways can the committee be formed?

We are tasked with forming a committee of 2 men and 3 women from a group of 4 boys and 5 girls. The number of ways to form such a committee involves calculating combinations.

### Step 1: Choose 2 boys from 4 boys
The number of ways to choose 2 boys from 4 can be calculated using the combination formula:
\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
For this, we have \( n = 4 \) and \( r = 2 \):
\[
\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
\]

### Step 2: Choose 3 girls from 5 girls
Similarly, the number of ways to choose 3 girls from 5 is:
\[
\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10
\]

### Step 3: Multiply the results
Since the choices of boys and girls are independent, the total number of ways to form the committee is the product of the two combinations:
\[
\text{Total number of ways} = \binom{4}{2} \times \binom{5}{3} = 6 \times 10 = 60
\]

Thus, the number of ways to form the committee is \( \boxed{60} \).

Would you like further details or have any questions?

Here are 5 related questions:
1. What if the committee requires 3 boys and 2 girls instead?
2. How many ways can a committee of 1 boy and 4 girls be formed?
3. How does the result change if you are allowed to pick fewer members for the committee?
4. What is the probability of selecting a specific boy and girl to be on the committee?
5. How many ways can you select all 9 people in the group?

**Tip:** Combination problems often involve the multiplication rule when choosing independently from multiple groups.

Learn how to solve a combinatorics problem involving selecting 2 boys from 4 and 3 girls from 5 to form a committee. This solution explores the use of the combination formula and the principle of counting. Suitable for students in Grades 10-12.

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