Selecting a committee of n people

[vishal.pathak]

Company X has n regional offices, where n represents an even integer. Each regional office must recommend two candidates, one male and one female, to serve on the corporate auditing committee. If each of the offices must be represented by exactly one member on the auditing committee and if the committee must consist of an equal number of male and female employees, how many different committees can be formed?
[spoiler]Soln. ( n!^2 )/ [ (n/2)!^4 ].
Company has n offices and from each office one male and one female will be recommended. Hence there will be n males and n females who will be recommended.
Each office must be represented by exactly one member which implies that totally only n people will be elected to the committee.
Also, there must be equal number of men and women which means that there must be n/2 men and n/2 women.
So out of n men, n/2 should be selected
and out of n women, n/2 should be selected.
Total ways of doing that
nC(n/2) x nC(n/2) = ( n!^2 )/ [ (n/2)!^4 ].

I have a doubt in this solution. When we do nC(n/2) x nC(n/2), are we not counting the ways in which we select both the man and the woman of the same regional office in the committee. Consider a regional office A. When we do nC(n/2) for n, we are randomly selecting people from the set of n and there will be a number of cases in which the guy from office A is slected. The same is true for the girl of office A. So when we do nC(n/2) x nC(n/2), will there not be a committee which will have both the guy and the girl of office A [/spoiler]

[user123321]

i am not so sure about this solution...this solution i think is not proper.
IMO it should be nC(n/2) = n!/(0.5n!)^2.

if we select half of the team teams and put men there, then it will give the combinations we need.

user123321

[vishal.pathak]

i am not so sure about this solution...this solution i think is not proper.
IMO it should be nC(n/2) = n!/(0.5n!)^2.

if we select half of the team teams and put men there, then it will give the combinations we need.

user123321

I dont believe this to be correct,buddy.We have to select n out of 2n and not n/2 out of n

Any other thoughts will be welcome

[user123321]

if you select n out of 2n you wont get sets with equal number of men and women.
instead, select half of teams and take a man from them & from remaining teams, take a woman. then it will solve the problem.

user123321

[Ian Stewart]

Company X has n regional offices, where n represents an even integer. Each regional office must recommend two candidates, one male and one female, to serve on the corporate auditing committee. If each of the offices must be represented by exactly one member on the auditing committee and if the committee must consist of an equal number of male and female employees, how many different committees can be formed?
Soln. ( n!^2 )/ [ (n/2)!^4 ].
Company has n offices and from each office one male and one female will be recommended. Hence there will be n males and n females who will be recommended.
Each office must be represented by exactly one member which implies that totally only n people will be elected to the committee.
Also, there must be equal number of men and women which means that there must be n/2 men and n/2 women.
So out of n men, n/2 should be selected
and out of n women, n/2 should be selected.
Total ways of doing that
nC(n/2) x nC(n/2) = ( n!^2 )/ [ (n/2)!^4 ].

I have a doubt in this solution. When we do nC(n/2) x nC(n/2), are we not counting the ways in which we select both the man and the woman of the same regional office in the committee. Consider a regional office A. When we do nC(n/2) for n, we are randomly selecting people from the set of n and there will be a number of cases in which the guy from office A is slected. The same is true for the girl of office A. So when we do nC(n/2) x nC(n/2), will there not be a committee which will have both the guy and the girl of office A

You're right - that solution is completely wrong, since it ignores the fact that we must choose one person from each office. Where is it from?

To pick a committee consisting of n/2 men and n/2 women, we just need to choose which half of the offices will contribute a woman to the committee; then the remaining offices will each be forced to contribute a man to the committee. From the n offices available, we can choose the n/2 offices who will send a woman to the committee in (n)C(n/2) ways, so that's the answer.

[knight247]

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