Suppose that a department contains 10 men and 17 women. How many ways are there to form a committee with 6 members if it must have strictly more women than men?
To choose 5 women and 1 man, there are
(17,5)*(10,1) ways.
Figure out the number of ways to choose
6 women,
4 women and 2 men,
and add up the total number of ways.
To find the number of ways to form a committee with 6 members, with more women than men, we can break it down into steps:
Step 1: Calculate the number of ways to choose different numbers of men and women for the committee.
We can have 1 man and 5 women, 2 men and 4 women, 3 men and 3 women, 4 men and 2 women, or 5 men and 1 woman.
Step 2: Calculate the number of ways to choose men and women for each scenario.
For 1 man and 5 women:
- Number of ways to choose 1 man from 10 men: C(10, 1) = 10
- Number of ways to choose 5 women from 17 women: C(17, 5) = 6188
For 2 men and 4 women:
- Number of ways to choose 2 men from 10 men: C(10, 2) = 45
- Number of ways to choose 4 women from 17 women: C(17, 4) = 2380
For 3 men and 3 women:
- Number of ways to choose 3 men from 10 men: C(10, 3) = 120
- Number of ways to choose 3 women from 17 women: C(17, 3) = 680
For 4 men and 2 women:
- Number of ways to choose 4 men from 10 men: C(10, 4) = 210
- Number of ways to choose 2 women from 17 women: C(17, 2) = 136
For 5 men and 1 woman:
- Number of ways to choose 5 men from 10 men: C(10, 5) = 252
- Number of ways to choose 1 woman from 17 women: C(17, 1) = 17
Step 3: Calculate the total number of ways to form the committee.
Add up the number of ways for each scenario: 10 * 6188 + 45 * 2380 + 120 * 680 + 210 * 136 + 252 * 17 = 904,360.
Therefore, there are a total of 904,360 ways to form a committee with 6 members, where the committee must have strictly more women than men.