Math Is Fun Forum

Thanks bob for your answer but I'm not sure I explained the problem right. For the permutations the "math is fun" calculator gave me 33. The 2 p's are not suppose to be distinguishable from each other. What the calculator did was list them first - which does indeed give you 60 then removed the duplicate permutations giving me 33. But I don't know how to do that with a formula.  Hopefully this will make things clearer:

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Permutations without repetition (n=5, r=3)
Using Items: a,p,p,l,e
Warning: your items have duplicates

List has 60 entries.
After removal of duplicates in result, there are now 33 entries.
{a,p,p} {a,p,l} {a,p,e} {a,l,p} {a,l,e} {a,e,p} {a,e,l} {p,a,p} {p,a,l} {p,a,e} {p,p,a} {p,p,l} {p,p,e} {p,l,a} {p,l,p} {p,l,e} {p,e,a} {p,e,p} {p,e,l} {l,a,p} {l,a,e} {l,p,a} {l,p,p} {l,p,e} {l,e,a} {l,e,p} {e,a,p} {e,a,l} {e,p,a} {e,p,p} {e,p,l} {e,l,a} {e,l,p}
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The above is how I am trying to list them. For a more generalized case I found this interesting problem on stack overflow but the explanation went a bit over my head. https://math.stackexchange.com/question … ion-subset His problem is similar to the one I posted but a bit more complex. We have the following letters: c,b,b,a,a,d and are making permutations of length 4. On that board they figured out the answer was that 102 distinct permutations could be made and indeed the "math is fun" calculator showed the same result.

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Permutations without repetition (n=6, r=4)
Using Items: c,b,b,a,a,d
Warning: your items have duplicates

List has 360 entries.
After removal of duplicates in result, there are now 102 entries.
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They showed a formula on stack overflow but as I said before I didn't understand it. Hopefully now I have made things a bit clearer.