# Math Help - problem

1. ## problem

A team of 7 people is to be chosen from 12
couples. How many ways can a team of three men and four
women be chosen? How many ways can a team containing
no couple be chosen? How many ways can a team
containing exactly two couples be chosen?

my guess is for part 1 is: (12 choose 7) * (6 choose 3) * (6 choose 4)?

cant figure out the rest!

2. Hello, dannyshox!

Your counting is off . . .

A team of 7 people is to be chosen from 12 couples.

(a) How many ways can a team of 3 men and 4 women be chosen?
There are 12 couples: 12 men and 12 women.

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

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

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

(b) How many ways can a team containing no couples be chosen?
I found no neat formula for this . . . I made an exhaustive list.

7 Men, 0 Women
. . . $n(\text{7M}) \;=\;{12\choose7}$ ways.

6 Men, 1 Woman
Choose 6 men: . ${12\choose6}$ ways.

The 1 woman must be chosen from the 6 women
. . who are not wives of the 6 men: . ${6\choose1}$ ways.
. . $n(\text{6M, 1W}) \:=\:{12\choose6}{6\choose1}$ ways.

5 Men, 2 Women
Choose 5 men: . ${12\choose5}$ ways.

The 2 women must be chosen from the 7 women
. . who are not wives of the 5 men: . ${7\choose2}$ ways.
. . $n(\text{5M, 2W}) \:=\:{12\choose5}{7\choose2}$ ways.

4 Men, 3 Women
Choose 4 men: . ${12\choose4}$ ways.

The 3 women must be chosen from the 8 women
. . who are not wives of the 4 men: . ${8\choose3}$ ways.
. . $n(\text{4M, 3W}) \:=\:{12\choose4}{8\choose3}$ ways.

3 Men, 4 Women
Choose 3 men: . ${12\choose3}$ ways.

The 4 women must be chosen from the 9 women
. . who are not wives of the 3 men: . ${9\choose4}$ ways.
. . $n(\text{3M, 4W}) \:=\:{12\choose3}{9\choose4}$ ways.

2 Men, 5 Women
Choose 2 men: . ${12\choose2}$ ways.

The 5 women must be chosen from the 10 women
. . who are not wives of the 2 men: . ${10\choose5}$ ways.
. . $n(\text{2M, 5W}) \:=\:{12\choose2}{10\choose5}$ ways.

1 Man, 6 Women
Choose 1 man: . ${12\choose1}$ ways.

The 6 women must be chosen from the 11 women
. . who are not wives of the 2 men: . ${11\choose6}$ ways.
. . $n(\text{1M, 6W}) \:=\:{12\choose1}{11\choose6}$ ways.

0 Man, 7 Women
Choose 7 women: . ${12\choose7}$ ways.

Now add them . . .

(c) How many ways can a team containing exactly two couples be chosen?

Choose 2 of the 12 men to be on the team: . ${12\choose2} \:=\:66$ 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: . ${4\choose3} \:=\:4$

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

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

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

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

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

3. Originally Posted by Soroban
Hello, dannyshox!

Your counting is off . . .

There are 12 couples: 12 men and 12 women.

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

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

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

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

7 Men, 0 Women
. . . $n(\text{7M}) \;=\;{12\choose7}$ ways.

6 Men, 1 Woman
Choose 6 men: . ${12\choose6}$ ways.

The 1 woman must be chosen from the 6 women
. . who are not wives of the 6 men: . ${6\choose1}$ ways.
. . $n(\text{6M, 1W}) \:=\:{12\choose6}{6\choose1}$ ways.

5 Men, 2 Women
Choose 5 men: . ${12\choose5}$ ways.

The 2 women must be chosen from the 7 women
. . who are not wives of the 5 men: . ${7\choose2}$ ways.
. . $n(\text{5M, 2W}) \:=\:{12\choose5}{7\choose2}$ ways.

4 Men, 3 Women
Choose 4 men: . ${12\choose4}$ ways.

The 3 women must be chosen from the 8 women
. . who are not wives of the 4 men: . ${8\choose3}$ ways.
. . $n(\text{4M, 3W}) \:=\:{12\choose4}{8\choose3}$ ways.

3 Men, 4 Women
Choose 3 men: . ${12\choose3}$ ways.

The 4 women must be chosen from the 9 women
. . who are not wives of the 3 men: . ${9\choose4}$ ways.
. . $n(\text{3M, 4W}) \:=\:{12\choose3}{9\choose4}$ ways.

2 Men, 5 Women
Choose 2 men: . ${12\choose2}$ ways.

The 5 women must be chosen from the 10 women
. . who are not wives of the 2 men: . ${10\choose5}$ ways.
. . $n(\text{2M, 5W}) \:=\:{12\choose2}{10\choose5}$ ways.

1 Man, 6 Women
Choose 1 man: . ${12\choose1}$ ways.

The 6 women must be chosen from the 11 women
. . who are not wives of the 2 men: . ${11\choose6}$ ways.
. . $n(\text{1M, 6W}) \:=\:{12\choose1}{11\choose6}$ ways.

0 Man, 7 Women
Choose 7 women: . ${12\choose7}$ ways.

Now add them . . .

Choose 2 of the 12 men to be on the team: . ${12\choose2} \:=\:66$ 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: . ${4\choose3} \:=\:4$

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

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

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

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

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

@Soroban - hi
For b)
An easy way - as there can't be a couple so first select 7couples from 12 couples. 12C7.
Now that you have 7 couples select either of the partner. i.e. 2^7ways.
So we have 12C7.2^7

solution to c) also appears wrong at following places plz

We must choose 3 more people from the other 4 men and 4 women
. . and include no couples.

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