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Thread: Combinatorics Question

  1. #1
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    Exclamation Combinatorics Question

    I'm unsure how to approach this problem. For a, is it just how many times there are (b, a). So say a = 2, and b = 1, then a forward path would be (1, 2) so there would only be (1, 1), (1, 2) so only 2 forward paths?

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  2. #2
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    Re: Combinatorics Question

    Quote Originally Posted by Psych View Post
    I'm unsure how to approach this problem. For a, is it just how many times there are (b, a). So say a = 2, and b = 1, then a forward path would be (1, 2) so there would only be (1, 1), (1, 2) so only 2 forward paths?
    I can give you another to look at these. Think of ordinary city-block grid if we move from $(0,0)\to(4,7)$ moving only one block east or one block north at a time. We must move four blocks to the east and seven blocks to the north. Here is a possible walk: $NNENENENENN$ That is a string of seven $N's$ for north & for $E's$ for east.
    There are $\dfrac{(7+4)!}{(7!)(4!)}=\dfrac{11!}{(7!)(4!)}$ ways to arrange that string an each arrangement is a possible path.

    My answer to your part a) is $\dfrac{(a+b)!}{(a!)(b!)}$

    Now you post the rest.
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    Re: Combinatorics Question

    So following that same logic for part b, it should be (c+d-a-b)!/((c-a)!(d-b)!) ways because it's just moving the origin correct?
    Last edited by Psych; Jan 21st 2019 at 07:36 PM.
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    Re: Combinatorics Question

    Quote Originally Posted by Psych View Post
    So following that same logic for part b, it should be (c+d-a-b)!/((c-a)!(d-b)!) ways because it's just moving the origin correct?
    Yes that is correct.
    For part c) $\displaystyle \frac{{\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)!}}{{\prod\limits_{k = 1}^n {\left[ {{(a_k)!}} \right]} }}$
    Last edited by Plato; Jan 22nd 2019 at 08:48 AM.
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