In a firm there are 20 men and 10 women. In how many can you have
a committee with 3 men and 2 women?

To solve this problem, we need to form a committee consisting of 3 men and 2 women from a group of 20 men and 10 women.

1. **Choosing 3 men out of 20**:  
   The number of ways to choose 3 men from 20 is given by the combination formula:
   \[
   \binom{n}{r} = \frac{n!}{r!(n - r)!}
   \]
   where \( n \) is the total number of men (20) and \( r \) is the number of men to choose (3). Thus, the number of ways to choose 3 men is:
   \[
   \binom{20}{3} = \frac{20!}{3!(20 - 3)!} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140
   \]

2. **Choosing 2 women out of 10**:  
   Similarly, the number of ways to choose 2 women from 10 is:
   \[
   \binom{10}{2} = \frac{10!}{2!(10 - 2)!} = \frac{10 \times 9}{2 \times 1} = 45
   \]

3. **Total number of ways to form the committee**:  
   Since the choices of men and women are independent, we multiply the two results:
   \[
   1140 \times 45 = 51300
   \]

Thus, the total number of ways to form a committee with 3 men and 2 women is **51,300**.

Would you like more details on this, or do you have any questions?

Here are 5 related questions to further your understanding:
1. How would the solution change if you needed a committee of 4 men and 1 woman?
2. What is the total number of ways to form any committee of 5 members from this group?
3. How would you calculate the probability of randomly selecting a committee with exactly 3 men and 2 women?
4. If the number of women increased to 15, how would that affect the number of possible committees?
5. What if the committee needed to have at least 1 woman—how would you approach that?

**Tip:** For problems involving selection or grouping, combinations (not permutations) are typically used, since the order of selection doesn’t matter.

In a firm there are 20 men and 10 women. In how many ways can you have a committee with 3 men and 2 women?

Learn how to calculate the number of ways to form a committee with 3 men and 2 women from a group of 20 men and 10 women. This solution involves combinatorics and the combination formula, suitable for students in Grades 10-12.

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