combinations: Find number of ways that mixed double game can be arranged amongst 9 married couples given that...

[tarun]

Q) Find the number of ways in which a mixed double game can be arranged from amongst 9 married couples given that no husband and wife play in the same game.

 

[JeffM]

How many ways can I choose two men from a set of nine?

Now, how many women do I have to choose from to make up the quartet of players. So how many ways can that choice be made.

Now that we have our four players, how many ways can I pair the two women with one of the men?

Now can you do this problem?

 

[tarun]

How many ways can I choose two men from a set of nine?

Now, how many women do I have to choose from to make up the quartet of players. So how many ways can that choice be made.

Now that we have our four players, how many ways can I pair the two women with one of the men?

Now can you do this problem?

thanks for the help.
i was able to get to the right answer which is (9c2)(7c2)×2
I tried to solve this in a different way by selecting one guy first and then another guy from the remaing 8 but i got the wrong answer - (9c1)(8c1)(8c1)(7c1) i think both are correct but obviously they are not , can you explain why picking one at a time wont work in this case?

 

[pka]

i was able to get to the right answer which is (9c2)(7c2)×2
I tried to solve this in a different way by selecting one guy first and then another guy from the remaing 8 but i got the wrong answer - (9c1)(8c1)(8c1)(7c1) i think both are correct but obviously they are not , can you explain why picking one at a time wont work in this case?

What you said was the correct answer, \(\displaystyle \dbinom{9}{2}\cdot 2\) answers only the number of ways to form only one doubles match.
It seems as if you want to form four. Is that correct?
Here is an interesting fact there \(\displaystyle 9\cdot\dfrac{8!}{2^4\cdot 4!}\) ways to group those nine women into four groups of two each. There are of course equally many ways to do the same for the men. Can you match a group of two men with a group two women so no pairs are a couple? I think that turns out to be a night-mare.

 

[JeffM]

I do not understand pka's answer so I may be wrong here.

My way gives the answer of

[MATH]\dbinom{9}{2} * \dbinom{7}{2} * 2 = \dfrac{9 * 8 * 7 * 6 * 2}{2 * 2} = 1512.[/MATH]
Your way of

[MATH]\dbinom{9}{1} * \dbinom{8}{1} * \dbinom{8}{1} * \dbinom{7}{1} = 4032[/MATH]
is far too large.

You can attack it your way, but it requires great care.

First to form the pair of men you can choose one man from 9 and then another man from the 8 remaining. But there are not now 8 woman to choose from: each man has a wife, reducing the number of woman to choose from to 7. So you can first choose one woman from 7 and then choose another woman from the remaining 6.

So your first mistake was there.

Your second mistake was not realizing that you counted each possible pair of men twice. You could pick A first and B second or vice versa. You need to divide by 2.

Similarly, with the woman, you could pick X first and then pick Y or pick Y first and then pick X. Again you need to divide by 2.

And given the male pair of A and B and the femaile pair of X and Y, we can form teams AX and BY or teams AY and BX. So you need to multiply by 2.

So if you do your method correctly you get

[MATH]\dfrac{\dbinom{9}{1} * \dbinom{8}{1}}{2} * \dfrac{\dbinom{7}{1} * \dbinom{6}{1}}{2} * 2 =[/MATH]
[MATH]\dfrac{9 * 8}{2} * \dfrac{7 * 6}{2} * 2 = 36 * 21 * 2 = 1512.[/MATH]
As I say, I may be spouting nonsense. Pka is very good at this stuff.

 

[pka]

Here is an interesting fact there \(\displaystyle 9\cdot\dfrac{8!}{2^4\cdot 4!}\) ways to group those nine women into four groups of two each. There are of course equally many ways to do the same for the men. Can you match a group of two men with a group two women so no pairs are a couple? I think that turns out to be a night-mare.

I do not understand pka's answer so I may be wrong here.

My way gives the answer of

[MATH]\dbinom{9}{2} * \dbinom{7}{2} * 2 = \dfrac{9 * 8 * 7 * 6 * 2}{2 * 2} = 1512.[/MATH]
Your way of

[MATH]\dbinom{9}{1} * \dbinom{8}{1} * \dbinom{8}{1} * \dbinom{7}{1} = 4032[/MATH]
is far too large.

You can attack it your way, but it requires great care.

First to form the pair of men you can choose one man from 9 and then another man from the 8 remaining. But there are not now 8 woman to choose from: each man has a wife, reducing the number of woman to choose from to 7. So you can first choose one woman from 7 and then choose another woman from the remaining 6.

So your first mistake was there.

Your second mistake was not realizing that you counted each possible pair of men twice. You could pick A first and B second or vice versa. You need to divide by 2.

Similarly, with the woman, you could pick X first and then pick Y or pick Y first and then pick X. Again you need to divide by 2.

And given the male pair of A and B and the femaile pair of X and Y, we can form teams AX and BY or teams AY and BX. So you need to multiply by 2.

So if you do your method correctly you get

[MATH]\dfrac{\dbinom{9}{1} * \dbinom{8}{1}}{2} * \dfrac{\dbinom{7}{1} * \dbinom{6}{1}}{2} * 2 =[/MATH]
[MATH]\dfrac{9 * 8}{2} * \dfrac{7 * 6}{2} * 2 = 36 * 21 * 2 = 1512.[/MATH]
As I say, I may be spouting nonsense. Pka is very good at this stuff.

I do not understand pka's answer so I may be wrong here.

My way gives the answer of

[MATH]\dbinom{9}{2} * \dbinom{7}{2} * 2 = \dfrac{9 * 8 * 7 * 6 * 2}{2 * 2} = 1512.[/MATH]

Sorry but it should be
What you said was the correct answer, \(\displaystyle \dbinom{9}{2}\dbinom{7}{2}\cdot 2\) answers only the number of ways to form only one doubles match.
Jeff is correct. Try to set just two matches.

 

[tarun]

thank you jeffM and pka.I totally understood where i was wrong.