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Say I have a set with the numbers 1, 2, 3... m.

How many unique combinations can be made of n numbers out of that set?

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  • $\begingroup$ This must be homework/basic combinatorics/book talks about this $\endgroup$
    – user5137
    Commented Dec 20, 2011 at 21:09
  • $\begingroup$ No, I am not a student. I just want to explain my theory in this answer. Sorry for asking this very simple question (for you folks), but for me this is the fastest way to get an answer (I think/hope). $\endgroup$
    – NGLN
    Commented Dec 20, 2011 at 21:10

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Here is a step-by-step approach:

Step 1: How many choices do you have for the first position?

Step 2: How many choices do you have for the second position?

...

Step $n$: How many choices do you have for the $n^\textrm{th}$ position?

Your answer should follow when you ponder the answers to the above questions.

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  • $\begingroup$ Ok, of course: (m-1+m-2+m-3+... for n=2). (Education is slightly coming back.) But how to enter that in a calculator or Excel? What is the name of such a calculation? It's not faculty. $\endgroup$
    – NGLN
    Commented Dec 20, 2011 at 21:19
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    $\begingroup$ If you view 123 as different from 132, you have permutations: en.wikipedia.org/wiki/Permutation. If you think they are the same (it depends on your application) you have combinations: en.wikipedia.org/wiki/Combination. The difference is permutations=combinations*n! to account for the different orders of n things. $\endgroup$ Commented Dec 20, 2011 at 21:23
  • $\begingroup$ @RossMillikan Omg, it really ís basic. I'm talking about combinations. n!/(k!(n-k)!) is what I was looking for. Thanks! $\endgroup$
    – NGLN
    Commented Dec 20, 2011 at 22:00
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    $\begingroup$ @NGLN: Glad to be of help. If you don't know the word, it can be hard to look up. And good for you for using enough parentheses to make it clear what is under the division slash. $\endgroup$ Commented Dec 21, 2011 at 0:22

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