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?

  • $\begingroup$ 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. $\endgroup$ – Christian Blatter Apr 21 '17 at 15:56

In this question nothing mentioned about their order of picking. So we don't need to think about that.

In case you need to arrange them in order. You can multiply with 6!


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