Asked by Fred
In how many ways can a comittee of 3 women and 4 men be chosen from 8 women and 7 men if two particular women refuse to serve on the committe together? Answer is 1750 but I have no clue how to get to it
Answers
Answered by
Reiny
Let's choose 3 women and 4 men from the 8 women and 7 men without any restrictions.
= C(8,3) x C(7,4) = ....
Now choose with the two particular women in place and th 4 men from the 7
That means we have to choose 1 more women from the remaining 6, the men situation does not change
= C(6,1) x C(7,4)
subtract this from the total found first, that should give you 1750
(worked for me)
= C(8,3) x C(7,4) = ....
Now choose with the two particular women in place and th 4 men from the 7
That means we have to choose 1 more women from the remaining 6, the men situation does not change
= C(6,1) x C(7,4)
subtract this from the total found first, that should give you 1750
(worked for me)
Answered by
scott
men ... 7C4 possibilities ... 35
women ... 8C3 possibilities ... 56
... but 6 of the groups can't be used ... the incompatible pair with any of the other 6 women
35 * (56 - 6) = ?
women ... 8C3 possibilities ... 56
... but 6 of the groups can't be used ... the incompatible pair with any of the other 6 women
35 * (56 - 6) = ?