Asked by Gilmore
Assume that you have a total of 13 people on the committee: 4 black males, 4 black females, 1 white male, and 4 white females. Party rules require that at least one black and at least one female hold one of the three offices.
In how many ways can the officers be chosen while still conforming to party rules?
Any ideas on how to solve this problem?
In how many ways can the officers be chosen while still conforming to party rules?
Any ideas on how to solve this problem?
Answers
Answered by
drwls
I would start with the number of ways of selecting three people with no restrictions, which is
13!/(3!*10!)= 286
and subtract the number of possible combinations that do not meet the requirements. For example, the number of combinations with no blacks (with only 5 nonblacks to choose among)is
5!/(3!*2!) = 10. Those combinations must be excluded. The number of ways of picking combinations with no females (with 5 males to choose among) is also 10.
That leaves 266
13!/(3!*10!)= 286
and subtract the number of possible combinations that do not meet the requirements. For example, the number of combinations with no blacks (with only 5 nonblacks to choose among)is
5!/(3!*2!) = 10. Those combinations must be excluded. The number of ways of picking combinations with no females (with 5 males to choose among) is also 10.
That leaves 266
Answered by
Dan Fox
P(13,3)=1716 (the total amount of combinations)
P(5,3)=60 (the total amount of non-blacks)
P(5,3)=60 (the total amount of non-females)
So subtract from the total the amount of combinations that won't work. So,
1716-60-60=1596
P(5,3)=60 (the total amount of non-blacks)
P(5,3)=60 (the total amount of non-females)
So subtract from the total the amount of combinations that won't work. So,
1716-60-60=1596