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Combinatorial Problem

  1. Feb 2, 2015 #1
    If we have the letters A, B, C, D, E, and F, and we are asked to find the number of arrangements where A is before B, wouldn't this just be half of the total number of arrangements with no restrictions? Intuitively this makes sense, but I have some doubts. For example, when I try to do the problem out, I get 2(5!)+2(4!)(2!)+(3!)(3!) which is slightly over half of the total arrangements with no restrictions.

    Help me out please!
  2. jcsd
  3. Feb 2, 2015 #2

    Simon Bridge

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    Try it for a smaller number of letters so that you can write out all the possibilities.
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