Suppose that a department contains 13 men and 19 women. How many ways are there to form a committee with 6 members if it must have strictly more women than men
Define n choose r as (n,r)=n!/(r!(n-r)!)
Number of ways to choose 2 men and 4 women
=(13,2)*(19,4)
Number of ways to choose 1 man and 5 women
=(13,1)*(19,5)
Number of ways to choose 0 man and 6 women
=(13,0)*(19,6)
Total=?
To find the number of ways to form a committee with 6 members, where there must be more women than men, we can use the concept of combinations.
First, let's consider the possible scenarios for the number of women on the committee.
1. If all 6 members are women: This situation is not possible since there are only 19 women available.
2. If 5 members are women and 1 member is a man: We can choose 5 women from the 19 available and 1 man from the 13 available. This can be done in C(19, 5) * C(13, 1) ways.
3. If 4 members are women and 2 members are men: We can choose 4 women from the 19 available and 2 men from the 13 available. This can be done in C(19, 4) * C(13, 2) ways.
4. If 3 members are women and 3 members are men: We can choose 3 women from the 19 available and 3 men from the 13 available. This can be done in C(19, 3) * C(13, 3) ways.
To find the total number of ways, we need to sum up all the possibilities: C(19, 5) * C(13, 1) + C(19, 4) * C(13, 2) + C(19, 3) * C(13, 3).
Evaluating this expression will give us the final answer.