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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$\begingroup$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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