sudhirmadaan wrote:

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.

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

_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372

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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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