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Thread: Combinatorics - choosing exactly k pairs from n

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    Combinatorics - choosing exactly k pairs from n

    Hi.
    I have the following combinatoric problem (well it's actually a probability problem that need to be resolved using combinatorics):

    There are $\displaystyle n$ pairs of shoes in the closet.
    $\displaystyle 2m$ shoes are chosen from it randomly. ($\displaystyle m<n$)
    find the probability to get exactly $\displaystyle k$ pairs.

    so this is what I'm thinking:

    first of all: $\displaystyle |\Omega |= \binom{2n}{2m}$ (the number of possibilities to choose 2m shoes from 2n).
    but I just can't decide how many possibilities are there to choose exactly $\displaystyle k$ pairs out of $\displaystyle n$.

    at first I thought it would be $\displaystyle \frac{\binom{n}{k}}{\binom{2n}{2m}}$, but that's not the right answer...
    any help would be greatly appretiated.

    *edit
    btw, acoording to what I found somwehre online, the answer is $\displaystyle \frac{\binom{n}{k}\binom{n-k}{2m-2k}2^{2m-2k}}{\binom{2n}{2m}}$
    I just couldn't figure out why...
    Last edited by Stormey; Nov 18th 2013 at 04:17 AM.
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    Re: Combinatorics - choosing exactly k pairs from n

    Quote Originally Posted by Stormey View Post
    There are $\displaystyle n$ pairs of shoes in the closet.
    $\displaystyle 2m$ shoes are chosen from it randomly. ($\displaystyle m<n$)
    find the probability to get exactly $\displaystyle k$ pairs.
    *edit
    btw, acoording to what I found somwehre online, the answer is $\displaystyle \frac{\binom{n}{k}\binom{n-k}{2m-2k}2^{2m-2k}}{\binom{2n}{2m}}$ I just couldn't figure out why...
    I have a minor disagreement with the online answer you found.
    I think it should be $\displaystyle \frac{\binom{n}{k}\binom{n-k}{m-k}2^{m-k}}{\binom{2n}{2m}}$

    Here are my reasons. Suppose we want to choose $\displaystyle J$ of those shoes so that none match (no pairs).
    Clearly $\displaystyle J\le n$. There are $\displaystyle \binom{n}{J}$ ways to choose $\displaystyle J$ pairs from the $\displaystyle n$ pairs
    From those $\displaystyle J$ pairs there are $\displaystyle 2^J$ ways to choose either the left or the right shoe from each pair.

    Does that help?
    Last edited by Plato; Nov 18th 2013 at 09:06 AM.
    Thanks from Stormey
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    Re: Combinatorics - choosing exactly k pairs from n

    Quote Originally Posted by Plato View Post
    Here are my reasons. Suppose we want to choose $\displaystyle J$ of those shoes so that none match (no pairs).
    Clearly $\displaystyle J\le n$. There are $\displaystyle \binom{n}{J}$ ways to choose $\displaystyle J$ pairs from the $\displaystyle n$ pairs
    From those $\displaystyle J$ pairs there are $\displaystyle 2^J$ ways to choose either the left or the right shoe from each pair.

    Does that help?
    Actually, that makes perfect sense to me, but I still don't see why is it $\displaystyle \frac{\binom{n}{k}\binom{n-k}{m-k}2^{m-k}}{\binom{2n}{2m}}$.

    *edit
    OK, now I see it.
    I didn't notice that we actually choose m pairs of shoes, not just k.
    Thank you very much my friend.
    that was very helpful.
    Last edited by Stormey; Nov 18th 2013 at 11:53 AM.
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