# Why cannot we use $P(n+r-1 , r)$ to calculate permutations when repetition is allowed?

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?

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.
• can you please make a little edit explaining why $n+r−1$ works when dealing with Combinations with Repetition ? – AmirWG Aug 3 at 19:30