combinations problem i just can't get my head around (1 Viewer)

[-MC]

10 couples are in a tennis club for mixed doubles, which pits one team of one man and one woman against another team of one man against one woman.
how many possible combinations for 4 people to play mixed doubles are possible if partners are allowed to play in the same game, but not on the same team?
If you only want the question, dont bother reading the rest

we're assuming all the couples are straight here, and the answer should come to 130.

so if i get 10C2 combinations for the first team, out of those 5 combinations will be invalid (5 couples)

then the second team is left to be made from the remaining 8 members. and this is where it all falls apart for me if one guy is chosen out of the remaining 4, and his wife is on the previous team, he can choose from 4 women, if his wife is NOT on team one, he can only choose from 3. I can't multiply these possibilities together because some possibilities will be counted twice.

what do i do?

 

[Carrotsticks]

10 couples or 10 people in the game?

 

[Carrotsticks]

I did it, and my answer was very different to 130. Are you sure that's correct? I can not see how it is so.

 

[badquinton304]

10 couples are in a tennis club for mixed doubles, which pits one team of one man and one woman against another team of one man against one woman.
how many possible combinations for 4 people to play mixed doubles are possible if partners are allowed to play in the same game, but not on the same team?
If you only want the question, dont bother reading the rest

we're assuming all the couples are straight here, and the answer should come to 130.

so if i get 10C2 combinations for the first team, out of those 5 combinations will be invalid (5 couples)

then the second team is left to be made from the remaining 8 members. and this is where it all falls apart for me if one guy is chosen out of the remaining 4, and his wife is on the previous team, he can choose from 4 women, if his wife is NOT on team one, he can only choose from 3. I can't multiply these possibilities together because some possibilities will be counted twice.

what do i do?

I havent done maths for a while but here goes.
Try to consider the men and women seperately.
10C2 assumes that any individual can be paired with another, whereas 5C1*5C1 assumes one woman is chosen and can be paired up with another man you would then take away 5 from this because of the five invalid combinations this leaves us with 20 possible teams.
Assuming one possible team combination is taken. We now have 4 men and 4 women. Assuming Mr X wife is in the team 1 and Mrs Y husband is in the team 1, then mr x is like tiger woods, he can go where he likes same for mrs y. Therefore we can say that for team 2, we have a 4C1*4C1 but instead of 4 invalid couples we have 3 because mr x and mrs y, not having a spouse present means that they can be together.Hence you would have 13 possible teams for the second team.
And thus 20 possible team 1 and 13 possible team 2 would give 13*20 or 260 possible groups of 4.
To visualise this problem particularly the team two part you should draw lines connecting each possible male female combination.

However I do not know if this is the correct solution, i may be wrong, do you have the answer.

 

[Rezen]

badquinton, wouldn't your solution count each pair of teams twice?

as in, your also ordering the pairs of team on the court, ie team A vs Team B is different to Team B vs Team A. to fix this devide 260 by two for the overcounting.

...And this leads to 130, the answer?

 

[badquinton304]

badquinton, wouldn't your solution count each pair of teams twice?

as in, your also ordering the pairs of team on the court, ie team A vs Team B is different to Team B vs Team A. to fix this devide 260 by two for the overcounting.

...And this leads to 130, the answer?

I suspected that but wasn't too sure, havent done this for a while. I am pretty sure the calculations for team 1 and team 2 are correct though, but I am not sure about how to bring them together.

 

[Rezen]

Yea, your calculations for team 1 and team 2 are correct but they also overcount.

*tries to explain it better* your method allows for the same two teams to be picked in different order. ie, your pick mrA and msB for the first team and then you pick mrC and ms D for the second. but you also countMrA and msB for the first and mrC and msD for the second, but this is the same as the first and hence you overcounted by 2.

But your calculations for the teams are essentially correct.

 

[nat_doc]

case one: 2 couples
10C2 (pick the couples).... and i think thats it cause they will be chosen so that H1W1 and H2W2 will be playing, but there is only one posibility that will be H1W2 and H2W1...
case 2: one couple
10C1 *18C2
case 3: no couoples:
20C4
ans:10C2+10C1*18C2+20C4

 

[random-1005]

20C1*19c1*18C1*17C1 *2

that is no where near right

 

[hscishard]

10 couples are in a tennis club for mixed doubles, which pits one team of one man and one woman against another team of one man against one woman.
how many possible combinations for 4 people to play mixed doubles are possible if partners are allowed to play in the same game, but not on the same team?
If you only want the question, dont bother reading the rest

we're assuming all the couples are straight here, and the answer should come to 130.

so if i get 10C2 combinations for the first team, out of those 5 combinations will be invalid (5 couples)

then the second team is left to be made from the remaining 8 members. and this is where it all falls apart for me if one guy is chosen out of the remaining 4, and his wife is on the previous team, he can choose from 4 women, if his wife is NOT on team one, he can only choose from 3. I can't multiply these possibilities together because some possibilities will be counted twice.

what do i do?

10 couples = 20 people

Oh right, LOL