how many committee of five people can be chosen from 20 men and 12 women if a) exactly three men must be on each committee b) at least four women on each committee
To solve this problem, we can use the combination formula:
nCr = n! / (r! * (n-r)!)
a) For exactly three men on each committee:
First, we need to select 3 men out of 20, which can be done in 20C3 ways.
Then, we need to select 2 women out of 12, which can be done in 12C2 ways.
The total number of committees can be computed as (20C3) * (12C2).
b) For at least four women on each committee:
In this case, we need to consider two scenarios:
- Four women and one man: We need to select 4 women out of 12, which can be done in 12C4 ways. Then we select 1 man out of 20, which can be done in 20C1 ways.
- Five women: We need to select 5 women out of 12, which can be done in 12C5 ways.
The total number of committees can be computed as (12C4) * (20C1) + (12C5).
Now let's calculate the values:
a) (20C3) = 20! / (3! * (20-3)!) = (20 * 19 * 18) / (3 * 2) = 1140
(12C2) = 12! / (2! * (12-2)!) = (12 * 11) / (2 * 1) = 66
Therefore, the total number of committees = 1140 * 66 = 75,240.
b) (12C4) = 12! / (4! * (12-4)!) = (12 * 11 * 10 * 9) / (4 * 3 * 2 * 1) = 495
(20C1) = 20! / (1! * (20-1)!) = 20
(12C5) = 12! / (5! * (12-5)!) = (12 * 11 * 10 * 9 * 8) / (5 * 4 * 3 * 2 * 1) = 792
Therefore, the total number of committees = 495 * 20 + 792 = 11,592.
Hence, there are 75,240 committees of five people for part (a) and 11,592 committees for part (b).