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Math Help - Number of ways to pair up group

  1. #1
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    Number of ways to pair up group

    Alright, I'm having a really hard time figuring this out even though I think it should be pretty basic. My question is:

    how many unique ways can you pair up a group 16 people so that each person is in a pair?

    Eg. if the question was how could you pair up 4 people (ABCD), the answer is 3. You can have AB/CD, AC/BD or AD/BC. Obviously CD/AB is the same as AB/CD so that "different" combination doesn't count. Brute force becomes pretty well impossible for 6 people or more though and I can't seem to google any formula which helps me out.
    any ideas?

    Thanks.
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  2. #2
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    Quote Originally Posted by Psychobabble View Post
    Alright, I'm having a really hard time figuring this out even though I think it should be pretty basic. My question is:

    how many unique ways can you pair up a group 16 people so that each person is in a pair?

    Eg. if the question was how could you pair up 4 people (ABCD), the answer is 3. You can have AB/CD, AC/BD or AD/BC. Obviously CD/AB is the same as AB/CD so that "different" combination doesn't count. Brute force becomes pretty well impossible for 6 people or more though and I can't seem to google any formula which helps me out.
    any ideas?

    Thanks.
    I believe it's just 15 * 13 * ... * 1. Reasoning: label them {A,B,...,P} and consider who is paired with A. There are 15 ways to do this. Then set that pair aside and relabel them {A,B,...,N} and consider how many ways there are to pair person A. Etc.
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  3. #3
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    Lazy man's "proof" for my previous post: A001147 on OEIS.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Lemmie take a crack at it.

    Note that you will have 8 pairs. So imagine that you have 8 slots, and in each slot, you will put a pair. Now, in the first slot, how many ways can you get a pair to put in that slot? {16 \choose 2} (from the 16 people, choose 2 to make a pair). For the second slot? {14 \choose 2} (choose a pair from the remaining 14 people). Continue this way until the seventh slot, where you will have 4 people left. In the seventh slot, you can put a pair in there {4 \choose 2} ways. At this point, you will only have a pair left. So you can put {2 \choose 2} = 1 pair in the last slot.

    But now, these pairs can be permuted, so divide by 8! Hence, the number of pairings are:

    \frac {{16 \choose 2}{14 \choose 2}{12 \choose 2} \cdots {2 \choose 2}}{8!}

    Which will give you the same number as the previous answer.
    Last edited by Jhevon; June 27th 2010 at 07:37 PM.
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  5. #5
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    wow, so if I calculate this correctly there's 2,027,025 possible ways to pair up 16 people? That's way higher than I was expecting.
    thanks for the quick responses guys!
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  6. #6
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    Quote Originally Posted by Psychobabble View Post
    wow, so if I calculate this correctly there's 2,027,025 possible ways to pair up 16 people? That's way higher than I was expecting.
    thanks for the quick responses guys!
    Yup, 2,027,025 ways :-)
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  7. #7
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    ah, now I see my problem and why I though it'd be much lower. I'm actually asking the wrong question.

    To be more precise, the question I'm asking is (and sorry for the frivolity here!) what are the chances of a tied score happening in my fantasy australian rules football league? there's 16 people in a league who are randomly paired up each week. So far there's been 10 weeks of competition and in that time teams have scored the same score in the same week four times which is the number I'll use as the chance that any two teams will score the same score in a particular week (this isn't exactly right, but it's the best way of doing it because calculating the objective chance is impossible given the number of variables). I was looking for the right denominator in this equation though, and 22m (the previous answer x10 weeks) is not that number because I was asking the wrong question.

    So the question I'm actually asking is something like - given a 40% chance that two teams will score the same points in any particular week, what is the chance that those two teams will be matched up against each other?

    Does that make sense? Sorry to lead people down the garden path with my first question!
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  8. #8
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    Quote Originally Posted by Psychobabble View Post
    ah, now I see my problem and why I though it'd be much lower. I'm actually asking the wrong question.

    To be more precise, the question I'm asking is (and sorry for the frivolity here!) what are the chances of a tied score happening in my fantasy australian rules football league? there's 16 people in a league who are randomly paired up each week. So far there's been 10 weeks of competition and in that time teams have scored the same score in the same week four times which is the number I'll use as the chance that any two teams will score the same score in a particular week (this isn't exactly right, but it's the best way of doing it because calculating the objective chance is impossible given the number of variables). I was looking for the right denominator in this equation though, and 22m (the previous answer x10 weeks) is not that number because I was asking the wrong question.

    So the question I'm actually asking is something like - given a 40% chance that two teams will score the same points in any particular week, what is the chance that those two teams will be matched up against each other?

    Does that make sense? Sorry to lead people down the garden path with my first question!
    To be honest I don't really understand your question. But if you're looking for the probability that two randomly selected teams are paired together, should be 1/15. (Because each team can be paired with one of 15 other teams, and each possibility is equally likely.)
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  9. #9
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    to state it in a slightly different way - assume is a 40% chance that two people a group of 16 people will be wearing a blue jumper. what is the probability of those two people being paired up if the group is randomly paired off?
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  10. #10
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    Quote Originally Posted by Psychobabble View Post
    to state it in a slightly different way - assume is a 40% chance that two people a group of 16 people will be wearing a blue jumper. what is the probability of those two people being paired up if the group is randomly paired off?
    Sorry but that didn't help. Maybe it's because I haven't played fantasy football. Question: When you wrote in post #7:

    Quote Originally Posted by Psychobabble View Post
    there's 16 people in a league who are randomly paired up each week.
    Did you mean 16 teams instead of 16 people?

    Okay so let me see if I'm getting this right though. You have 16 people, label them {A,B,...,P}. You've preselected two of them, say, A and B. There is a 40% chance that A and B will both wear blue jumpers. You are given that all 16 people will be paired randomly. What is the chance that A and B will be paired together? Since the color of their jumpers is not a factor in how they are paired, the answer is just 1/15.

    So as you can see, I still don't know what you're asking.
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  11. #11
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    that's my answer then . There's a 40% chance the "blue jumper" condition will be satisfied for any group of 16, and a 1/15 chance they'll be paired together. (4/10)*(1/15)=2.667%.

    Thanks for making me see that the answer's really obvious .
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