# Ways in which a mixed double game can be arranged from amongest $5$ married couples

The number of ways in which a mixed double game can be arranged from amongest $5$ married

couples if at least one husband and wife play in the same game.

My Try::

no. of ways in which least one husband and wife play in the same game = Total - no. of ways in which no. husband and wife play in same game.

for Total first we will select $4$ players out of Total $10$ players is $= \displaystyle \binom{10}{4}$

Now We have calculate no. husband and wife play in same game

First we will select $2-$ pairs out of $5$ which can be done by $\displaystyle \binom{5}{2}$ like $H_{1},H_{2}$ and $W_{1},W_{2}$

So we have two possibility which is $\left\{\left(H_{1},W_{2}\right)\;,\left(H_{2},W_{1}\right)\right\}$

So answer is $= 190$ but answer given is $= 140$

can anyone explain me where i am wrong. Thanks

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Mixed doubles means that each team of $2$ consists of a man and a woman.

We can choose $2$ men and $2$ women in $\binom{5}{2}\binom{5}{2}$ ways.

Now we count the number of choices of $2$ men and $2$ women in which there is no "couple." The women can be chosen in $\binom{5}{2}$ ways. For each of these ways, the men can only be chosen in $\binom{3}{2}$ ways, for a total of $\binom{5}{2}\binom{3}{2}$.

Calculate and subtract. We get $70$.

Now for every "good" choice of $4$ people, we can divide these into a pair of man/woman teams in $2$ ways. So there is a total of $140$ possibilities.

Another way: We can also do the calculation by counting directly the number of "good" choices of $2$ men and $2$ women.

We could choose $2$ couples. This can be done in $\binom{5}{2}=10$ ways.

Or we can choose $1$ couple and two people, one of each gender, who are not a couple. The couple can be chosen in $5$ ways. The other man can then be chosen in $4$ ways, and for each choice the other woman can be chosen in $3$ ways, for a total of $60$.

Add our $10$ two couple choices. We get $70$. Then double as in the first solution.

Remark: The incorrect calculation in the OP may be largely due to lack of knowledge of tennis terminology.

Starting with $\binom{10}{4}$ is not a good idea. For then we have to take away the all female choices, the $3$ female $1$ male choices, and so on. Then we have to remove the no couple choices. More work.

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Thanks André Nicolas Yes I have no knowledge of tennis terminology –  juantheron Feb 9 '13 at 7:01

Out of 5 couples we have to select 2 persons from men and 2 from woman so that same couple doesn't select. this can be done as follows,1st select 2 men from 5 i.e,$(5,2)$ and and select 2 woman from the remaining 3, (since 2 husbands selected of two wifes exclude that). So no of selections are $(5,2)*(3,2)=30$. These can be done in two ways so final answer is $2 \times 30=60$.

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