Question 199754: May someone can help me with:
In how many ways can 2 men, 4 women, 3 boys, 3 girls be selected from 6 men, 8 women, 4 boys and 5 girls if a particular man and a particular woman must be selected?
The answer is 7000 but I get 14000:
{5!/(2!3!)} x {7!/(4!3!)} x {4!/3!} x {5!/(2!3!)}
for the men, I take 5! because there are (6-1) men remaining: one has been selected already;
for the women, I take 7! because there are (8-1) women remaining: one has been selected already;
Thanks a lot!
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
You have one very small error. Actually, you made the same error twice, but one of the instances did not affect the outcome of the calculation.
You were absolutely correct to assume that, because one of the men was already chosen, that you were selecting from a group of 5 rather than 6. Likewise, since a woman was pre-selected, you are selecting from 7 rather than 8.
The problem is, in the case of your calculation for the number of combinations for the men, not only are you selecting from a group that has been reduced by 1, the number you need to select is also reduced by 1.
Hence, the combination calculation for the men must be:
!} = 5)
It appears that you made the same error when you did the combinations calculation for the women, but here there is no impact on the calculation because:
!} = \frac{7!}{4!(7-4)!} =\ _7C_4)
So you can see that the error didn't make any difference at that point.
Bottom line:
John
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