## how many ways-married couples

For the discussion of math. Duh.

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evinda
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### how many ways-married couples

Hello!!!
Could you help me at the following exercise?
With how many ways can we choose a man and a woman that are not married to each other from n married couples?

I thought that it is (n-1)n ,but I am not sure.Can you tell me if it is right?

Flumble
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### Re: how many ways-married couples

Can you explain how you got to (n-1)n? Or was it simply a hunch?

Since #(choose unmarried man and woman) = #(choose man and woman) - #(choose married man and woman), I'd approach it from the right-hand side:
Spoiler:
there are n! ways to match the men to the women (a.k.a. there are n! permutations to sort the women keeping the men in place; you don't get different couples by shuffling the men) and there are n married couples, so there ought to be n!-n ways to pick a man and a woman who aren't married.

On a sidenote: must all couples consist of a man and a woman?

 As gmalivuk pointed out I assumed all ways to pair up everyone instead of the ways to choose one couple.
Spoiler:
Which is quite easy, as you just have to choose one man and a woman he didn't marry (or vice-versa, doesn't matter), which makes for n*(n-1) possibilities.
 Or put differently: there are n*n ways to pick a male and a female, but n of those are the original couples.
Last edited by Flumble on Fri Apr 25, 2014 7:14 pm UTC, edited 3 times in total.

gmalivuk
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### Re: how many ways-married couples

You seem to be counting the total possible ways to pair up all the men with all the women (excluding cases where a man is paired with his wife). The OP seems to be asking for a count of individual man+woman pairs.

Though you're right that it is complicated by the fact that they haven't told us how many of the couples are gay.
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DavCrav
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### Re: how many ways-married couples

evinda wrote:Hello!!!
Could you help me at the following exercise?
With how many ways can we choose a man and a woman that are not married to each other from n married couples?

I thought that it is (n-1)n ,but I am not sure.Can you tell me if it is right?

So firstly, how many ways are there of choosing a man and a woman? How many of those choices lead to a married couple? Subtract one from the other.

Thesh
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### Re: how many ways-married couples

So is the question intended to be basically equivalent to the following?

You have n pegs and n holes, each peg fits into only one of the holes. How many ways can you place the pegs next to the holes so that no peg is next to a hole it fits into?
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somehow
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### Re: how many ways-married couples

No, I don't think that's equivalent. I think the OP is asking how many ways we can pick one peg and one hole such that the peg does not fit into the hole.

Spoiler:
Each man is a member of n-1 possible pairs consisting of a man and a woman not married to each other (one for each of the other n-1 couples besides his own). There are n men. This gives n*(n-1) = n2-n possible choices of one man and one woman such that they are not married. Each such pair contains exactly one of the n men and so will have been counted exactly once.
Xenomortis wrote:O(n2) takes on new meaning when trying to find pairs of socks in the morning.

patzer
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### Re: how many ways-married couples

Not enough information. How many gay married couples exist out of the total? Are all of the people either men or women, or are there some who are neither or both? How many of the people are polygamous?

Assuming entirely heterosexual monogamous couples:

Spoiler:
There are n men and n women. For each of the n men there are n-1 women who are not his wife. So the total number is n(n-1), or n2-n.

Assuming heterosexual and homosexual monogamous couples:

Spoiler:
Assume there are a male-male couples, b female-female couples, and c male-female couples. (a+b+c=n). There are 2a+c men and 2b+c women.

There are (2a+c)(2b+c) ways to pair a male with a female; with c of these pairings being subtracted for being a heterosexual couple. For a total of 4ab2+2ac+2bc+c2-c.

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ConMan
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### Re: how many ways-married couples

If we're looking for ways to pair up men and women from a set of heterosexual monogamous couples such that no-one ends up with their spouse, then we're asking for the number of derangements of the set of women (or equivalently men). As for the number of ways to choose a man and then choose a woman who is not his wife, then it is n*(n-1).
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