A committee of 8 persons is to be formed from 10 men and 10 women. How many different committees are possible if the committee must contain at least two women and at least one of the two oldest men?

To solve this problem, we need to consider different scenarios separately and then add the results together.

1. First, let's consider the scenario where the committee includes one of the two oldest men. There are 2 ways to choose this man. Then, we need to choose 7 more members from the remaining 18 people (10 women and 8 men). This can be done in C(18, 7) ways, where C(n, r) denotes the number of combinations of n items taken r at a time.

2. Next, let's consider the scenario where the committee includes both of the two oldest men. There is only 1 way to choose these two men. Then, we need to choose 6 more members from the remaining 18 people. This can be done in C(18, 6) ways.

However, in both scenarios, we need to subtract the cases where there are less than two women in the committee.

In the first scenario, there are C(8, 7) ways to choose 7 men from the remaining 8 men, and 10 ways to choose 1 woman.

In the second scenario, there are C(8, 6) ways to choose 6 men from the remaining 8 men, and 10 ways to choose 1 woman, and C(10, 2) ways to choose 2 women.

So, the total number of different committees is:

2C(18, 7) + C(18, 6) - [10C(8, 7) + C(8, 6)*10 + C(8, 6)*C(10, 2)].

You can calculate the values of C(n, r) using the formula C(n, r) = n! / [r!(n-r)!], where n! denotes the factorial of n, which is the product of all positive integers up to n.