Exercise on Combinatorics (combinations)

In the book A First Course in probability (8th edition) there is a problem at the end on the paragraph about combinatorics which states:

From a group of 8 women and 6 men a committee consisting of 3 men and 3 women is to be formed. How many different committees are possible if: a) 2 of the men refuse to serve together?

I guessed I could find all the possible committee of 3 men and 3 women, then subtract the number of committees which contains the 2 men. Thus:

$${8 \choose 3}{6 \choose 3} - 4{8 \choose 3} = 896$$

Which is correct, according to the solutions provided at the end of the book.

But later i thought, isn't 896 the number of committe of ordered couple of a group of 3 men and a group of 3 women? Isn't, say:

{Jim, Jay, John} {Martha, Annah, Stacy}

different from

{Martha, Annah, Stacy} {Jim, Jay John}, by this formula?

Am I "ordering" the 2 groups by not dividing by 2 each of them?

• Your formula is correct. It does not envisage the sixtuple (Martha, Annah, Stacy, Jim, Jay John) at all, but only counts sixtuples beginning with three men. – Christian Blatter Apr 21 '17 at 15:56