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
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