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"

3Comments
Abaa @wakawaka
25 Feb 2017 08:38 pm

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

shivani @shivani1234
9 Oct 2017 12:04 pm

@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

ANUMOLU SUNIL @sunilanumolu
30 Aug 2018 09:09 pm
In the numerator also you have to divide by 2! beacuse there also you will get the same groups twice