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.