Tricky perms and combs questio (1 Viewer)

Thjv · Oct 23, 2015

In a tennis club, there are five married couples available to play a "mixed doubles" match, that is, a match in which a combination of one man and one woman play against a combination of another man and another woman. In how many ways can a group of four persons be chosen for this match if a man and his wife play in a match but not as partners.

InteGrand · Oct 23, 2015

Thjv said:
In a tennis club, there are five married couples available to play a "mixed doubles" match, that is, a match in which a combination of one man and one woman play against a combination of another man and another woman. In how many ways can a group of four persons be chosen for this match if a man and his wife play in a match but not as partners.

$Let's treat the two sides of the tennis court (say the north side and south side) as different. We'll divide by $2$ at the end to make up for this.$

$First, we'll choose the pair for the north side. We can pick the male player in $5$ ways and his partner in $4$ ways (as the doubles partner can't be the spouse). So this is $5\times 4 = 20$ ways.$

$Now, for the south side, there are precisely two cases for picking the male:$

$Case 1: The male we pick has his wife chosen for the north side. In this case, there is just 1 way to choose the male, and 4 ways to choose the female (as we can pick any of the remaining females), so there are $1\times 4 = 4$ ways.$

$Case 2: The male we pick does \textsl{not} have his wife chosen for the north side. In this case, we can pick any male (so 4 ways), and there are 3 ways to then pick the female (as his wife is in the list of those we can pick from, and we can't pick her), for a total of $4\times 3 = 12$ ways for Case 2.$

$Adding the numbers for Cases 1 and 2, the no. of ways to pick the south side players is $4+12 = 16$.$

$Multiplying the no. of ways to pick the team for the north side by that for the south side gives $20 \times 16 = 320$.$

$Dividing by two since the sides (north and south) are actually irrelevant, the answer is $320\div 2 = 160$.$

InteGrand · Oct 23, 2015

godofindolence · Oct 23, 2015

Hey, not sure if I'm reading the question right, but shouldn't the answer be 80

considering that it is mixed doubles, wouldn't the maximum possibility be 5C2*5C2 = 100

I think the stipulation states that one specific man has to be in the game with his wife (uh I think this is wrong, but I'm assuming only one pair of couples have to be in the game), but not as partners, so the possibilities would be
M (5)* W (4) * M (4) * W (1) <-- this has to be the wife of one of the 5 men which is 80.

calamebe · Oct 23, 2015

InteGrand said:
$Let's treat the two sides of the tennis court (say the north side and south side) as different. We'll divide by $2$ at the end to make up for this.$

$First, we'll choose the pair for the north side. We can pick the male player in $5$ ways and his partner in $4$ ways (as the doubles partner can't be the spouse). So this is $5\times 4 = 20$ ways.$

$Now, for the south side, there are precisely two cases for picking the male:$

$Case 1: The male we pick has his wife chosen for the north side. In this case, there is just 1 way to choose the male, and 4 ways to choose the female (as we can pick any of the remaining females), so there are $1\times 4 = 4$ ways.$

$Case 2: The male we pick does \textsl{not} have his wife chosen for the north side. In this case, we can pick any male (so 4 ways), and there are 3 ways to then pick the female (as his wife is in the list of those we can pick from, and we can't pick her), for a total of $4\times 3 = 12$ ways for Case 2.$

$Adding the numbers for Cases 1 and 2, the no. of ways to pick the south side players is $4+12 = 16$.$

$Multiplying the no. of ways to pick the team for the north side by that for the south side gives $20 \times 16 = 320$.$

$Dividing by two since the sides (north and south) are actually irrelevant, the answer is $320\div 2 = 160$.$

Hey, I noticed that (5x4x4x4)/2 also results with 160. Is this just lucky, or is it also a valid way to do the question?
I did for the top side, 5x4, and bottom 4x4 (I knew there were two cases just couldn't figure out how many combinations they were so just did 4x4) and divided by 2 (as the sides are the same).

InteGrand · Oct 23, 2015

calamebe said:
Hey, I noticed that (5x4x4x4)/2 also results with 160. Is this just lucky, or is it also a valid way to do the question?
I did for the top side, 5x4, and bottom 4x4 (I knew there were two cases just couldn't figure out how many combinations they were so just did 4x4) and divided by 2 (as the sides are the same).

$Yeah it's just luck that the guess of $4\times 4$ happens to be equal to the answer obtained by Cases 1 and 2 ($1\times 4 + 4\times 3$).$

InteGrand · Oct 23, 2015

godofindolence said:
considering that it is mixed doubles, wouldn't the maximum possibility be 5C2*5C2 = 100

No, because to choose a pair, we don't choose 2 from 5. We choose 1 from 5 and one from another 5, which is more than doing ⁵C₂.

InteGrand · Oct 23, 2015

godofindolence said:
Hey, not sure if I'm reading the question right, but shouldn't the answer be 80

considering that it is mixed doubles, wouldn't the maximum possibility be 5C2*5C2 = 100

I think the stipulation states that one specific man has to be in the game with his wife (uh I think this is wrong, but I'm assuming only one pair of couples have to be in the game), but not as partners, so the possibilities would be
M (5)* W (4) * M (4) * W (1) <-- this has to be the wife of one of the 5 men which is 80.

The question wording doesn't mean that a specific man or pair has to be in the game I think; I think it's just asking for the number of ways to choose people for a match if we need mixed doubles pairs and spouses can't be on the same team.

godofindolence · Oct 23, 2015

InteGrand said:
No, because to choose a pair, we don't choose 2 from 5. We choose 1 from 5 and one from another 5, which is more than doing ⁵C₂.

Ahh I see, thanks! So it would be 5*5*4*4

InteGrand · Oct 23, 2015

godofindolence said:
Ahh I see, thanks! So it would be 5*5*4*4

$It's not that, because this also counts cases where spouses are paired up with each other, which we said we cannot have.$

godofindolence · Oct 23, 2015

InteGrand said:
$It's not that, because this also counts cases where spouses are paired up with each other, which we said we cannot have.$

Ah yes, it's just the maximum possible mixed double pairings without restrictions.
Ok I see where I went wrong now, I should have used permutations so 5P2*5P2

InteGrand · Oct 23, 2015

godofindolence said:
Ah yes, it's just the maximum possible mixed double pairings without restrictions.

$Actually we would need to divide that answer by $2! = 2$, since the two pairs are not ordered. I.e. $5\times 5 \times 4 \times 4$ considers these two matchups as \emph{different} matchups:$

$$\bullet$ (Alex, Alice) vs. (Ben, Cindy)$

$and$

$$\bullet$ (Ben, Cindy) vs. (Alex, Alice).$

$$Clearly those two pairs of teams are same for us, since the order is unimportant. But this is not accounted for in $5\times 5 \times 4 \times 4$, so we need to divide by $2! = 2$.$

godofindolence · Oct 23, 2015

Ah yes!
Ughhh, I hate permutations and combinations, everything makes sense and seems like its right until you realise it's wrong.

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