##### 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?

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"

A = Total ways of making 2 groups of size 6 = C(12, 6) * 1

B= Ways of having (3 men and 3 women in group 1) & (3 men and 3 women in group 2) = C(6,3)*C(6,3)*1*1

Here Ways of having 3 men in group 1 = Ways of having 3 women in group 1 = C(6,3). After that only 1 way of creating group-2.

Ans = B/A

@vignesh see to @wakawaka solution, it is correct , there is no need of dividing it by 2! and men and women are differentiable so how could you say that group is unlabelled