Originally Posted by

**Soroban** Hello, dannyshox!

Your counting is off . . .

There are 12 couples: 12 men and 12 women.

Choose 3 men: .$\displaystyle {12\choose3} \:=\:\frac{12!}{3!\,9!} \:=\:220$ ways.

Choose 4 women: .$\displaystyle {12\choose4} \:=\:\frac{12!}{4!\,8!} \:=\:495$ ways.

Therefore: .$\displaystyle n(\text{3M, 4W}) \:=\:220\cdot 495 \:=\:108,\!900$ ways.

I found no neat formula for this . . . I made an exhaustive list.

**7 Men, 0 Women**

. . . $\displaystyle n(\text{7M}) \;=\;{12\choose7}$ ways.

**6 Men, 1 Woman**

Choose 6 men: .$\displaystyle {12\choose6}$ ways.

The 1 woman must be chosen from the 6 women

. . who are *not* wives of the 6 men: .$\displaystyle {6\choose1}$ ways.

. . $\displaystyle n(\text{6M, 1W}) \:=\:{12\choose6}{6\choose1}$ ways.

**5 Men, 2 Women**

Choose 5 men: .$\displaystyle {12\choose5}$ ways.

The 2 women must be chosen from the 7 women

. . who are *not* wives of the 5 men: .$\displaystyle {7\choose2}$ ways.

. . $\displaystyle n(\text{5M, 2W}) \:=\:{12\choose5}{7\choose2}$ ways.

**4 Men, 3 Women**

Choose 4 men: .$\displaystyle {12\choose4}$ ways.

The 3 women must be chosen from the 8 women

. . who are *not* wives of the 4 men: .$\displaystyle {8\choose3}$ ways.

. . $\displaystyle n(\text{4M, 3W}) \:=\:{12\choose4}{8\choose3}$ ways.

**3 Men, 4 Women**

Choose 3 men: .$\displaystyle {12\choose3}$ ways.

The 4 women must be chosen from the 9 women

. . who are *not* wives of the 3 men: .$\displaystyle {9\choose4}$ ways.

. . $\displaystyle n(\text{3M, 4W}) \:=\:{12\choose3}{9\choose4}$ ways.

**2 Men, 5 Women**

Choose 2 men: .$\displaystyle {12\choose2}$ ways.

The 5 women must be chosen from the 10 women

. . who are *not* wives of the 2 men: .$\displaystyle {10\choose5}$ ways.

. . $\displaystyle n(\text{2M, 5W}) \:=\:{12\choose2}{10\choose5}$ ways.

**1 Man, 6 Women**

Choose 1 man: .$\displaystyle {12\choose1}$ ways.

The 6 women must be chosen from the 11 women

. . who are *not* wives of the 2 men: .$\displaystyle {11\choose6}$ ways.

. . $\displaystyle n(\text{1M, 6W}) \:=\:{12\choose1}{11\choose6}$ ways.

**0 Man, 7 Women**

Choose 7 women: .$\displaystyle {12\choose7}$ ways.

Now add them . . .

Choose 2 of the 12 men to be on the team: .$\displaystyle {12\choose2} \:=\:66$ ways.

There is 1 way that their wives are also on the team.

We have 4 people on the team.: 2M, 2W.

We must choose 3 more people from the other 4 men and 4 women

. . and include *no* couples.

There are 4 cases:

. . **3M, 0W**: .$\displaystyle {4\choose3} \:=\:4$

. . **2M, 1W**: .$\displaystyle {4\choose2}{2\choose1} \:=\:12$

. . **1M, 3W**: .$\displaystyle {4\choose1}{3\choose2} \:=\:12$

. . **0M, 3W**: .$\displaystyle {4\choose3} \:=\:4$

There are: .$\displaystyle 4 + 12 + 12 + 4 \:=\:32$ ways.

Therefore: .$\displaystyle n(\text{exactly 2 couples}) \;=\;66\cdot32 \;=\;2112$ ways.