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]
[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]
