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A group of 6 men and 6 women is randomly
divided into 2 groups of size 6 each. What is the
probability that both groups will have the same
number of men?

ANSWER I AM GETTING IS  :       [C(6,3) * C(6,3)] / [12!/(6!*6!*2!)]
ANSWER GIVEN IS                :       [C(6,3) * C(6,3)] / [12!/(6!*6!)]
source : sheldon m ross textbook

please do verify and provide explanation ....
what i feel is that "the answer should be divided by 2! because the permutations among 2 groups shouldnot happen because both the groups are unlabelled"
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Please kindly refer d link mentioned below Yeah!!! See it's a case of Symmetric Probability Space. 

Exercise 49 mentioned in d Pdf mentioned below might just solve your doubt. :)

https://math.la.asu.edu/~jtaylor/teaching/Fall2010/STP421/lectures/lecture5.pdf

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