Permutation/Combination Problem

Question

I'll post the question, and then my answer in the following post. If someone could explain why my logic is wrong, I would appreciate it.

A group of three men and three women are asked to get in line, with the men in spots 1, 2 and 3 and the women in spots 4, 5 and 6.

How many different arangements are possible?

abmyers · Answer

I thought, ok, order does not matter, so combination not permutation.

Then I thought, 6!/(3!*3!) = 20

This is incorrect. Correct answer is 3! * 3!

I get why 3! * 3! is right, but I'm not sure I understand why my way is wrong. Can someone point out the error in my logic?

mainhoon · Answer

I think the formula you are using N!/M! P! is when we have N total and M similar and P similar objects - clearly that is not the case here. We are just ordering the first 3 and then the next 3 so that results in 3! x 3!. Now if you are thinking why cant I see this problem as diving up a group of 6 into 2 groups of 3 each - this has been discussed extensively by Bunuel and others - that is not this case either - this is only a subset of that larger problem and in that case the answer would be 6!/(2! 3^2)..

oldstudent · Answer

The formula 6!/(3!*3!) = 20 cannot be used here because we are talking about different individuals. say 3 mens have name M1, M2, M3 and women are W1,W2,W3 now in the line the arrangment M1 M2 M3 W1 W2 W3 is different than 
M1 M3 M2 W1 W2 W3 as M2 and M3 are different individuals. That is why as these arrangements are different we cannot divide it by 3!*3!.

while if we get a problem to find all possible arrangements of word MADAM - then it would be 5!/2!*2! simply bcos M and M are same and A and A are same.

Hope this makes sense to you!

aim730 · Answer

These are mutually exclusive ones Ok Tell me how many ways 3 men can be placed in position and 3 women be placed in 3 positions think about it and you will get your answer See the question ask about the above scenario. I hope it helps

Permutation/Combination Problem

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