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    I'm finding combinatorics to be a real pain in the backside at the moment! Even though the questions I'm trying to do are relatively simple, I'm not having problems with actually doing the maths but it's more about devising the problem in the first place.

    There are some things I would like verified:

    • The origin of the number of integer solutions to x_{1} + x_{2} + \ldots x_{r} = n, both positive and non-negative. According to my textbook these are ^{n-1}C_{r-1} and ^{n+r-1}C_{r-1} respectively but I don't understand why they hold. I thought about putting n objects into r boxes and finding the possible ways of arranging the 'dividers' ('|') and the 'objects' ('O') (e.g. OOO | O | OOOO | OO) but I don't know how to deduce the formulae above - I just know that there are (n + r - 1)! ways of arranging the objects and the dividers but I don't understand how to account for when the dividers are next to each other or at the ends (i.e. when we require positive integer solutions to the equation and we need no dividers next to each other).
    • Gaining a better understanding of the reasoning needed, because in some situations I try drawing a diagram to help me to visualise the ideas further, but in some problems that is a very long-winded approach and it turns out that the reasoning was simple, but I wouldn't have thought to use that particular reasoning. (e.g. if you've got the letters ABCDEF, how many ways are there where A occurs before B and B occurs before C, how would I have immediately known to use the fact that there are 3! = 6 of ordering A, B and C, and if there are 6! ways of arranging A, B, C, D, E, F, then there are 6!/3! = 120 ways in which A comes before B and B comes before C).
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    (Original post by OL1V3R)
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    It's easiest to look at the non-negative one first of all.

    You have r-1 dividers and n objects to put in.

    So total number is n+r-1, hence your (n+r-1)! arrangements, but the objects are all the same and so are the dividers, so divide by n! and by (r-1)!, which gives you the desired result.

    For the positive one.

    Pre-allocate a single object to each box. Then you are left with n-r objects to allocate to r boxes, and are allowed empty slots (as there is an object already pre-allocated there), and this is similar to the previous case - have a think on the details.


    And yeah, finding the right way to look at the problem is the key issue with these. Haven't got any tips I'm afraid.
 
 
 
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Updated: April 6, 2011
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