Playing mixed doubles

[Motherjane]

Help with this question please!!


A group of 4 couples want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

[chidcguy]

Is it mixed doubles or only doubles?

The answer will vary depending on that.

Your Q does not have the mixed doubles in it. Your post subject has mixed doubles in there.

If mixed doubles

Four teams can be formed with one M & F.

First team can be picked in 4 X 4 ways. Second team in 3 X 3, Third team in 2 X 2 ways and last team in 1 X 1 ways.

so thats 16 X 9 X 4 X 1 = 144 X 4 = 576 ways.

If Not mixed doubles

The first two out of 8 can be picked in 8 C 2 ways

The next two in 6 C 2 ways

The next two in 4 C 2 ways

The last two in 2 C 2 ways

8 C 2 X 6 C 2 X 4 C 2 X 2 C 2

[Ian Stewart]

The answer above is almost right. The teams aren't in any order (there is no 1st team, 2nd team, etc) so you also need to divide by 4!.

[chidcguy]

Ian,

I am not sure I understood your reason behind dividing with 4!. Can you be please kind enough to explain a bit more?

Thanks

[anayak]

Out of 8, first he can select 2 by 8C2 ways. Then from the remaining 6 he can select 2 in 6C2 ways. Similarly in total he can select in

8C2 * 6C2 * 4C2 * 2C2= 28*15*6*1= 2520 ways

[Motherjane]

Thanks Ian and chidcguy!!

Indeed the answer needs to be divided by 4!. I was thinking on it and reasoned that when the first team can be formed in 8C2 ways, since there are no overlap of players till now selected in the team, you will divide it by 1. Similarly when the second team will be formed in 6C2 ways, there might be overlap of players from the already selected 1st team as there is no restriction on the player to be selected in the second team. Hence divide it by the number of teams formed ie, 2. And so on .

Hence, (8C2 / 1) * (6C2 / 2) * ( 4C2 / 3) * (2C2 / 4)

Is the reasoning correct?

[Motherjane]

One more way.

Fix first person.So he/she can pair up with any 1 of the remaining 7 ppl.
Now, when 6 ppl are left,fix up smbdy in the second team.So he/she can now pair up with 1 of the remaining 5 ppl...

2 teams are formed and 4 ppl are left.
Again fix up somebody in the 3rd team.He/she can pair up with 1 of the remaining 3 ppl.

2 ppl are left .They can only form 1 team.

Hence, total ways to form a team = 7 * 5* 3 = 105 ...Ans

Hope this is correct.

[beeparoo]

Thanks Ian and chidcguy!!

Indeed the answer needs to be divided by 4!. I was thinking on it and reasoned that when the first team can be formed in 8C2 ways, since there are no overlap of players till now selected in the team, you will divide it by 1. Similarly when the second team will be formed in 6C2 ways, there might be overlap of players from the already selected 1st team as there is no restriction on the player to be selected in the second team. Hence divide it by the number of teams formed ie, 2. And so on .

Hence, (8C2 / 1) * (6C2 / 2) * ( 4C2 / 3) * (2C2 / 4)

Is the reasoning correct?

Hey Motherjane,

I'm puzzled by your choice of words when you describe the "overlap" of players from the 1st team into the 2nd team, and so forth. So, I have to ask-- is that a valid way of describing the logic?

For me, that kind of terminology is reserved for sets that may share common elements. For instance, I think of Venn diagrams and, say, surveys of people who are married/unmarried and their preferences for boxers/briefs.

Here, my initial thought was that the "overlap" logic was misapplied. I mean, once a person is selected into a team, the subsequent teams are created from the remaining players. So indeed, the limitation is defined...

I think your calculation of

(8C2 / 1) * (6C2 / 2) * ( 4C2 / 3) * (2C2 / 4)

is correct because you are eliminating the order of how players are arranged within each team.

Correct me if I'm wrong or half-wrong on this... I normally have difficulty with the theoretical concepts on complex combination/permutation problems.

Cheers,
Sandra

[Motherjane]

Hey Sandra,

I apologize for the word 'overlap' that I had used in my explanation.
I meant exactly what you have written down.

When I have considered the different pairings for each of the 4 teams , there are multiple repetitions.
Since there are 4 teams,4! teams have been repeated.

What I meant was that there is no limitation on a particular person to be only in 1 team and not in other. Its only that for one such combination when A is in team 1, then the rest of the 6 people can form the other 3 teams. But there is no limitation for A to be in only team 1.

[AleksandrM]

I am about as lost as a little boy in the woods.

[shankyrams]

I think the answer is 105.
If there are X*Y people who are put in X groups containing Y people each the this can be done in [ (X*Y) ! / ((X!)*(Y^X)) ] ways.

[gabriel]

Is the confusion over why is the division with 4! is necessary or is it over how we divide the groups. I (or the someone else) can answer accordingly.

[g_beatthegmat]

still confused

[Motherjane]

At first, I tried with 2 couples.
Suppose A1A2 & B1B2 are 2 couples.

So if A1 forms a team with any other person, there is only one way in which the other team would be formed.
Like A1 A2 -----> then only B1B2 will be the other team.

Similarly for A1B1 and A1B2 where the other team would be A2B2 and A2B1 respectively.

So a total of 3 ways the teams could be formed.

Hence,total eventual possibilities is total pairs possible / total no of teams.

Total pairs possible is 4C2 and total no. of teams possible will be only 2.
So total eventualities = 4C2/2 = 3.

Extending the logic above for 4 couples:

Suppose A1 A2, B1 B2 , C1 C2 and D1D2 are 4 couples.

The first two people among the group can be chosen in 8C2 ways. That forms the first team. The other 6 people in whatever way they arrange themselves will form 3 teams and not more.

Hence, the total arrangements will be : 8C2/4 (as there are 4 teams)

Now the second team will be formed in 6C2 ways. Whatever may be the permutation among the rest of the 4 people, the teams formed will be 2.

Hence, the total arrangements will be : 6C2/3 (as there are 3 teams)

The logic goes on for the next two team formation too.

Hence the total no. of ways will be:
(8C2/4) * (6C2/3) * (4C2/2) * (2C2/1) = 105.

Hope I could clear some confusion

[Ian Stewart]

Let's look at two closely related problems:

You have 8 people, and want to divide them into 4 teams of 2. One team will go to the Olympics, one to Wimbledon, one to the US Open and one to Australia. Notice the teams are in order now. We then have 8C2 choices for the Olympic team, 6C2 choices for the Wimbledon team, 4C2 choices for the US Open, and 2C2 = 1 choice left for the Australia team. In total you have 8C2*6C2*4C2*2C2 choices of teams. We do not divide by 4! at the end, because the order of the teams matters.

Now, let's look at the problem posted above. How many ways can 8 people be divided into 4 teams of 2? Notice in this version of the question, the order of the teams is not important. This question is different from the Olympics/Wimbledon/US Open/Australia question. Why? If I have 8 people, A, B, C, D, E, F, G and H, and send them to different tennis tournaments, then this set of teams:

Olympics: {A, B}
Wimbledon: {C, D}
US Open: {E, F}
Australia: {G, H}

is different from this set:

Olympics: {C, D}
Wimbledon: {A, B}
US Open: {E, F}
Australia: {G, H}

because I have a different team going to the Olympics in each case.

But if I'm dividing people up into teams, and the order of the teams doesn't matter, then {A, B}, {C, D}, {E, F}, {G, H} is the same set of teams as {C, D}, {A, B], {E, F}, {G, H}. These answers are different when the order matters, but are the same when the order does not matter. Because the order of 4 teams doesn't matter, we can count by:

a) pretending the order does matter;
b) dividing by 4! (the number of ways to arrange 4 things in order) because the order of 4 things doesn't matter.

So the answer must be:

(8C2*6C2*4C2*2C2)/4!

which is equal to 105.