# Choosing 4 men from 6, and 4 women from 7, with a restriction.

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