Probability of only one gender in the six group

Question

A total of twelve attendees consisting of six couples are attending a marriage seminar. An activity during the seminar had to randomly group the attendees into six group of two attendees. What is the probability that each group contains attendees of the same gender only ?

Answer 1

First let's develop the numerator of our probability fraction. 

 

Begin by considering the first pair which must be same gender; there are 6C2 ways to do this (from the pool of men, say). The next pair (of men) can be chosen from the 4 remaining men, so that's 4C2; the last two men can be chosen in 2C2 ways.  All 3 of these "nCr" numbers are to be multiplied (i.e. are factors appearing in the numerator). (It makes no difference what order these factors are in--it's just a matter of one gender first, the other next.]

 

Continuing our development of the numerator, we now have the 6 women remaining, and the number of ways to establish the various possible pairs would be the product 6C2•4C2•2C2 so the resulting completed numerator is

[6C2•4C2•2C2]².

 

What is the denominator? A set of 6 numbers, each representing the 'fulfillment' of a pair, regardless of gender composition: 12C2•10C2•8C2•6C2•4C2•2C2.

 

The quotient of numerator & denominator, described above, is the desired probability.

 

Frankly, this was a problem of greater than usual difficulty, so I am open to qualified critiques.  Thanks.