Company X has 6 regional offices. Each regional office must recommend

chetan2u Abhishek009, please tell me what m i missing here
Each office has two candidates M and F , so this way we got 6 M and 6 F
now question says only 1 M or F will be represent his/her office.
so my understanding says if M is selected than his counterpart F will not, and vice versa.
So 6M and 6F
so selection will be 6c3*3c3
where 6c3 represents 3 males out of 6 and 3c3 represents remaining 3 females , which are not male counterpart.
similarly we have case for female 6c3 * 3c3
so ultimatly we 2 6c3= 40.
Please correct me where I am wrong.

the committee must consist of 6 members. Further, because the committee must have an equal number of male and female employees, it must include 3 men and 3 women. First, let's form the female group of the committee. There are 3 women to be selected from 6 female candidates (one per region). One possible team selection can be represented as follows, where A, B, C, D, E, & F represent the 6 female candidates:
A B C D E F
Yes Yes Yes No No No
In the representation above, women A, B, and C are on the committee, while women D, E, and F are not. There are many other possible 3 women teams. Using the combination formula, the number of different combinations of three female committee members is 6! / (3! × 3!) = 720/36 = 20. To ensure that each region is represented by exactly one candidate, the group of men must be selected from the remaining three regions that are not represented by female employees. In other words, three of the regions have been “used up” in our selection of the female candidates. Since we have only 3 male candidates remaining (one for each of the three remaining regions), there is only one possible combination of 3 male employees for the committee. Thus, we have 20 possible groups of three females and 1 possible group of three males for a total of 20 × 1 = 20 possible groups of six committee members.

if each office can recommend 2 members, one male and one female, the potential members of the committee are 12, 6 males and 6 females. Since each office will be represented by 1 person and the committee must have an equal number of males and females, the committee will be formed by exactly 6 people, 3 males and 3 females.
Now we have to select 3 males in a group of 6 males and 3 females in a group of 6 females. Therefore, we use the combination formula: 6!/(3!*3!) -> number of ways of choosing either 3 males out of 6, or 3 females out of 6; and then this result must be squared, to multiply all the possible combinations of males and females.
Finally, the result is: 400

Total cases: 2^6=64
Not cases: 2+12+30=44
6M=1
6F=1
5M1F=6!/5!=6
5F1M=6!/5!=6
4M2F=6!/4!2!=15
4F2M=6!/4!2!=15
Favorable cases: 64-44=20

Ans (D)

There are 6 offices. You can list out the 6 offices as A through F.

And each office must recommend 1 male and 1 female. From each office exactly 1 person will go on the committee.

This means the committee will have 6 people.

Further, we are told that there must be an equal number of males and females on the committee. Therefore, we know that there must be 3 males and 3 females chosen on the 6 person committee.


The 12 recommendations will consist of the following:

Office A - Male A and Fem A

Office B - Male B and Fem B

….

Office F - Male F and Fem F


Since each office will send one person ——-> and the order in which the people are selected does not matter (only the makeup of the group selected is important)

all we care about is which 3 offices will send their male representative and which 3 offices will send their female representatives.


Out of the 6 offices, how many different ways can we have 3 offices send their male representative?

6 c 3 = 20 ways

For each one of these 20 ways, because each office must send exactly 1 rep and there must be 3 males and 3 females, the remaining 3 offices NOT chosen must send their Female rep


20 different ways

D

Posted from my mobile device

Best and simplest explanation. The question asks "how many such different committees can be formed" which means it is simply an arrangement question. Most are finding ways to arrange which is not what is asked. After 3 males and 3 females are are chosen we have to find the number of different committees we can form out of them i.e. the number of arrangements we can have which is 6! / 3! * 3! = 20. Thanks

hi chetan2u

I have tried to understand the solution provided by you but I cannot.

I am requesting you to explain why 6c3 * 6c3 would not work here.

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