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A team of 4 men and 4 women has to be chosen from 6 men and 7 women, so total combinations:

$^6C_{4} \times ^7C_{4} = 525$

If a restriction is placed that some two women will always be chosen together and ommitted together, what is the number of combinations now?

my first thought is grouping the two women together as a single block, so number of women can be regarded as 6; hence the combinations are then:

$^6C_{4} \times ^6C_{4} = 225$

Although this very answer is given at the back of my textbook, I doubt that it is correct, since this block of two women and other three women will make a team of 5 women, whereas a team of 4 is required. Am i wrong?

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I hope that the answer at the back didn't write ${}^6\text{C}_4$ for the second term ($15$, which is numerically equal to ${}^6\text{C}_4$, is fine). You can check that if for example there were $8$ women, the term for two specific women both in or both out would not be equal to ${}^7\text{C}_4$. –  André Nicolas Aug 22 '11 at 14:02

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up vote 2 down vote accepted

Yes - and, no. You can choose 4 women from the other 5 in 5-choose-4 (which is 5) ways, or take the block and 2 of the other women in 5-choose-2 (which is 10) ways, so all up there are 15 ways to choose the women. Well, 6-choose-4 is also 15, so it's the right answer, though maybe for the wrong reason!

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logic! so then does it make difference that 4men and 4women will make 4 man-woman pairs? –  schooler Aug 22 '11 at 15:37
    
@schooler, it doesn't make any difference to me. It might make a difference to them, I don't know. But perhaps I'm missing the point of your question. –  Gerry Myerson Aug 22 '11 at 22:51
    
how about 6C3-5C4 ? is that correct? –  schooler Aug 25 '11 at 12:24
    
@schooler, it gives the right number, but whether it gives it for the right reason depends on how you got it. There are a lot of ways to get the number 15, and some make sense while some are lucky coincidences. But maybe I don't understand what you are asking. –  Gerry Myerson Aug 25 '11 at 13:13

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