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What are Subset Pairs as mentioned in this problem?

  1. Oct 28, 2006 #1
    What are "Subset Pairs" as mentioned in this problem?

    In reference to this problem. BTW - I'm not asking for a solution or even a method of attack for a solution since I just enjoy doing these puzzles. I'm just trying to get clarification on one of the terms used.

    http://mathschallenge.net/index.php?section=project&ref=problems&id=106

    Say, for A={3, 5, 6, 7} (example pulled from here), what are the 25 subset pairs?
     
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  3. Oct 28, 2006 #2

    mathman

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    Something doesn't look right. There are only six pairs.
     
  4. Oct 28, 2006 #3

    Office_Shredder

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    Which problem is that from?
     
  5. Oct 28, 2006 #4
    I assume it means you can pair a 1 element set with a 1, 2, or 3 element set, but you'd definitely not have to compare 1x1 element sets or nxm element sets where n doesn't equal m. But still, there are (2^4-1) non-empty subsets in that case. :dunno: I just can't find a pairing that counts to 25.

    :uhh: I posted the links to it right there
     
    Last edited: Oct 28, 2006
  6. Oct 29, 2006 #5

    HallsofIvy

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    I think mathman was assuming that "subset pairs" meant "subsets with exactly two members"- there are only 6 of those.

    I, on the other hand, assumed it meant "pairs of subsets". Of course, since a 4 element set has 24= 16 subsets, there would be 16C2= 16!/(2! 14!)= 8(15)= 120 of those, not 25!
     
  7. Oct 29, 2006 #6

    shmoe

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    Both your links were to the project euler homepage, not to the specific problem you are looking at, which is possibly 106:

    http://mathschallenge.net/index.php?section=project&ref=view&id=106

    A subset pair in that problem is a pair of disjoint, non-empty subsets. For your example A={3,5,6,7}. Just sort them by the sizes, for example pairs where both sets have 1 element:

    {{3},{5}}, {{3},{6}}, {{3},{7}}, {{5},{6}}, {{5},{7}}, {{6},{7}}

    Then look at 2 and 1 elements, 3 and 1 elements, finally 2 and 2 elements.

    To count them, count ordered subset pairs instead, then divide by 2. There are 4 ways to pick the first subset to have size 1, then 2^(4-1)-1=7 ways to pick the second set in the pair, so 28 so far. There are 4!/2!/2!=6 ways to pick the first subset to have size 2, 2^(4-2)-1=3 ways to pick the second set, so 18 more. There are 4 ways to pick the first subset to have size 3, and 1 way to pick it's pair, so another 4, for a total of 50. Divide by 2 to kill off our over counting from considering the order as distinct.
     
  8. Oct 29, 2006 #7
    Maybe the links are different when you're logged in, but they go to 106 and 103 for me. :dunno:

    Thanks for that explanation, I found my error:

    What I finally had was:

    (4 choose 1)*(3 choose 1) + (4 choose 2)*(2 choose 2) + (4 choose 1)*(3 choose 2) + (4 choose 1)*(3 choose 3)

    I forgot to divide the first and second terms by two :)
     
  9. Oct 29, 2006 #8

    shmoe

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    That must be, looking at the url of your links they are slightly different than the one I used. Your version must send you to a default page if you're not signed in to that website.
     
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