The post above has the exact answer for the same.

If you have to place 3 identical red balls and 3 identical blue balls in 6 slots, you will take \(\frac{6!}{3!3!}\) as it is the formula to arrange 6 things out of which 3 are of one kind and the other 3 are of the other kind.
SO this will include mmffmf or mmmfff etc, and will not differentiate amongst males and amongst females.
But we are looking at mfmfmf only, and also all 3 males are taken as separate identity
here m have specific places and f have specific places. They can be arranged within these places, so 3!*3!

So does this mean in "GMAT language" that every 1st, 3rd and 5th person in the lineup MUST be a male? If so, I get it. I guess it's about practicing GMAT language...[/quote]
Hi pkloeti

We are going with easy working.

As highlighted male, female, male, female, male, female = male(1), female(1), male(2), female(2), male(3), female(3)

Now male(1) probability = 3 and female(1) probability = 3 (as total 3 males and 3 females)

male(2) probability = 2 and female(2) probability = 2 (because 1 male and 1 female are already at position 1. So remaining are 2)

In same way male(3) probability = 1 and female(3) probability = 1

Coming to highlighted part
male(1), female(1), male(2), female(2), male(3), female(3) = 3 * 3 * 2 * 2 * 1 * 1
= 36 is our answer.
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Hard to interpret this question. I thought they meant that the "MFMFMF" is one of how many permutations of 3 males and 3 females, where all males and all females are considered identical within a gender.

Some possibilities:
MMMFFF
MMFMFF
MMFFMF
MMFFFM
...

So 6! / (3! * 3!) = 20

The question very directly states that team A decides to lineup male, female, male, female, male, female. There is really only 1 way to interpret this.

This is more a verbal question than quant.