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by stillcrazy on July 12th, 2004

stillcrazy

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Is there a method to calculate the number of possible combinations in a selection from a group containing repeats?

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  • by n_o_u_s on September 2nd, 2005

    n_o_u_s

    This might be helpful:

    To count the number of ways to put n objects into two groups of size x and size y where x+y=n:

    n!/(x!*y!) = number of combinations. aka (n choose x). or (n choose y) for that matter. This useful when you want to choose a group of x items from n items because the others necessarily fall into a second group, the group you want to ignore - the y group.

    For some cases, number of arrangements with repeats = (n choose x) + (1 for each object that repeats once) + (3 for each object for each object that repeats twice). By 'repeats once,' I don't mean that it only gets to repeat once in the list of combinations, but rather that it is allowed once copy, so you might have AACB and AABD and AABC. In the work I did on paper, I denoted copies by having a bar over it. For those that allowed two copies, I use the letter with a ~ over it. I haven't worked out exactly the limitations of this method. I'm hoping that when I have more time for this, I can show that additions of 'k for each object that repeats j times' will follow a regular progression.

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