To solve this problem, we need to count the number of ways to form a committee of 8 people from a group of 15 men and 10 women, while satisfying the conditions:

1. At least 5 men in the committee, and
2. At least 2 women in the committee.

Let's break down the solution into cases based on the number of men and women in the committee:

Case 1: 5 Men and 3 Women

• Choose 5 men from 15: $\binom{15}{5}$
• Choose 3 women from 10: $\binom{10}{3}$

The total number of ways for this case is: $\binom{15}{5} \times \binom{10}{3} = \frac{15!}{5!(15-5)!} \times \frac{10!}{3!(10-3)!}$

Case 2: 6 Men and 2 Women

• Choose 6 men from 15: $\binom{15}{6}$
• Choose 2 women from 10: $\binom{10}{2}$

The total number of ways for this case is: $\binom{15}{6} \times \binom{10}{2} = \frac{15!}{6!(15-6)!} \times \frac{10!}{2!(10-2)!}$

Case 3: 7 Men and 1 Woman

• Choose 7 men from 15: $\binom{15}{7}$
• Choose 1 woman from 10: $\binom{10}{1}$

The total number of ways for this case is: $\binom{15}{7} \times \binom{10}{1} = \frac{15!}{7!(15-7)!} \times \frac{10!}{1!(10-1)!}$

Summing the Results

The total number of ways to form the committee, satisfying all conditions, is the sum of the results from the three cases: $\text{Total} = \binom{15}{5} \times \binom{10}{3} + \binom{15}{6} \times \binom{10}{2} + \binom{15}{7} \times \binom{10}{1}$

Now, we will calculate each term:

1. $\binom{15}{5} = 3003$ and $\binom{10}{3} = 120$, so $3003 \times 120 = 360360$.
2. $\binom{15}{6} = 5005$ and $\binom{10}{2} = 45$, so $5005 \times 45 = 225225$.
3. $\binom{15}{7} = 6435$ and $\binom{10}{1} = 10$, so $6435 \times 10 = 64350$.

Adding these together gives: $360360 + 225225 + 64350 = 650935$

Thus, the total number of ways to form the committee is 650,935.

Would you like a breakdown of the calculations or have any questions?

Related Questions:

1. How would the solution change if the committee had to have exactly 5 men?
2. What if there had to be at least 6 women on the committee?
3. How many ways are there to form a committee with no restrictions on gender?
4. How does the problem change if the total committee size is 10 instead of 8?
5. What if the group consisted of 20 men and 5 women?

Tip:

Use binomial coefficients to efficiently solve problems involving selections or combinations, especially when restrictions are present.