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

Question

Company X has 6 regional offices. 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?

A. 5
B. 10
C. 15
D. 20
E. 40

A. 5
B. 10
C. 15
D. 20
E. 40

vnigam21 · Accepted Answer

This is an easy question. I tried again and I was able to solve. :lol:

As the question says, If each of the offices must be represented by exactly one member on the auditing committee and there are 6 Regional Offices, so each of the regional offices will be represented by 6 members on the auditing committee. Also, as per question, committee must consist of an equal number of male and female employees that is 3 Male members and 3 Female members. So, the question is now simply reduced to the possible number of arrangements for MMMBBB . It can be done in 6!/3!3! ways that is equal to 20.

chetan2u · Answer

Hi,

the method or formula you are working on, is not correct here...

when you make a group of 3 from 6, the other group is automatically made as only 3 are left after choosing initial 3, and the same will be applicable here ..

Ways to divide 2 groups of 3 each from 6 members.... 
say ABCDEF are 6 persons and you choose ABC in 6C3, another group DEF is automaticaly formed..
so when you choose DEF in 6C3, ABC is also formed and thus there is a repetition and that is why divison by 2!..
ans 6C3/2!.....
But as you have found 2 * 6C3, which ACTUALLY should be 2*\frac{6C3}{2!} = 6C3. ....

SOLUTION : Another way :-

Let the 6 companies be ABCDEF..
Now the 3 we choose in 6C3 send male and the remaining 3 will send female....
so we basically find HOW many ways 2 teams of 3 each can be choosen out of 6 = 6!/3!3!2!= 10..

But each team will once send MALE and other time FEMALE ...
so ans = 10*2 = 20

sudhirmadaan · Answer

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.

mdbofficial · Answer

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

Vibhav10 · Answer

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.

exc4libur · Answer

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)

Fdambro294 · Answer

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

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RahulHGGmat · Answer

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.

shashwat25 · Answer

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

99percnetile · Answer

Hey - So the idea here is that when 3 males are selected from 6 offices the other offices are bound to send 3 Females . It does not matter which three females they send. If it were two be squared it would mean that each of the three males on the committee are being paired with specific 3 females on the committee resulting in 400 arrangements . Arrangements is not our concern here.

shaurya_gmat · Answer

6 Female Candidates, 6 Male Candidates -
Ways to chose 6 female candidates - 6C3 = 20
This leaves us with 3 offices to represent 3 males, i.e., 3C3 = 1

Therefore total = 20
Can someone check if this is a correct approach ?

tabishjaved · Answer

Step1: Choose 3M among these 6 packs = 6C3
3F will automatically be chosen from the remaining 3packs after Step1
Ans.= 6C3 = 20

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

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

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