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I know we can simply get permutations when repetition is allowed using $$n^r$$ But why cannot we use the normal permutation formula: $P (n, r) $ (repetition is not allowed) but with A little tweak: $$P (n+r-1, r) $$ For example, when $n=3$ and $r=2$ Using first way: $$3^2=9$$ Then I tried to use the second way: $P (3 + 2 - 1 , 2) = P (4 , 2) = 12$ (does not give the same results) $$$$ I saw this $n + r - 1$ thing in a formula which is used to calculate combinations when repetition is allowed: $C (n+r-1, r) $ and I was wondering why I cannot do the same with permutations?

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The reason $n^r$ works is that you have $n$ choices for each position. $P(n+r-1)=(n+r-1)(n+r-2)\ldots(n)$ would suggest you have $n+r-1$ choices for the first position. The number of choices for the first position cannot depend on the number of positions there are.

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  • $\begingroup$ can you please make a little edit explaining why $n+r−1$ works when dealing with Combinations with Repetition ? $\endgroup$ – AmirWG Aug 3 at 19:30
  • $\begingroup$ @AmirWG See this explanation for the number of combinations with repetition. $\endgroup$ – N. F. Taussig Aug 3 at 19:37

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