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"