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Math Help - Permutation and combination Q

  1. #1
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    Permutation and combination Q

    how many rectangles can you form by joining the squares of a chess board
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  2. #2
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    if you connect any 4 squares on a chessboard you will get a rectangle.

    There are 64 squares on a chessboard.


    Can you finish from there?
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    Quote Originally Posted by ice_syncer View Post
    how many rectangles can you form by joining the squares of a chess board
    I believe you're seeking the answer in this thread. (Scroll down a bit since there was a little issue with formatting to begin with. In particular, look at awkward's post.)

    SpringFan25's response in this thread is also good, if I'm interpreting properly. You would have to keep in mind that the four squares you connect are not necessarily distinct, and must be arranged such that the top two squares define a row, as do the bottom two squares, and same with left and right defining columns.

    Then again, maybe SpringFan25 was suggesting \binom{64}{4} and there is some interpretation issue going on.
    Last edited by undefined; June 19th 2010 at 11:35 AM.
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    so is the answer 64C4 ??
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    no its not, i wasn't paying attention sorry.

    Assuming it still counts as a rectangle if its 1 chess square wide, You can get every possible rectangle by choosing 2 (distinct) squares to be the opposite diagonal edges.

    there are 64C2 ways to do that (i think).

    Add 64 if the 1x1 "rectangle" counts as a solution.
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    The correct answer is \binom{9}{2}^2=1296.
    Look at the diagram. There nine vertical lattice points on the left edge and nine horizontal lattice points on the bottom edge. By choosing two points from each set we get a rectangle. (Yes squares are rectangles.)
    Attached Thumbnails Attached Thumbnails Permutation and combination Q-untitled.gif  
    Last edited by Plato; June 19th 2010 at 12:31 PM.
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    google agrees with plato. I remain mystified as to why my answer is wrong
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  8. #8
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    Quote Originally Posted by SpringFan25 View Post
    google agrees with plato. I remain mystified as to why my answer is wrong
    Your answer counts several duplicates. For example, choosing a8 and b7 is the same as choosing b8 and a7 (using Algebraic chess notation). Your method can be adjusted as follows.

    16\cdot\binom{8}{2}+\frac{\binom{64}{2}-16\cdot\binom{8}{2}}{2}+64=1296
    Last edited by undefined; June 19th 2010 at 11:34 AM.
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    I see latex error
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  10. #10
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by ice_syncer View Post
    I see latex error
    Not everything works quite right after the forum upgrade. I just clicked Edit Post followed by Save and it shows up properly now.

    Also, if you want to see what people typed, you can quote them.
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