Combinations vs permutations with replacement

combinationscombinatoricspermutations

The formulas for combinations and permutations with replacement are, respectively:

$$
C = \binom{n + k – 1}{k} \\
P = n^k
$$

Where $n$ and $k$ correspond to the number of distinct objects we are to draw $k$ times from with replacement.

My question is I can't wrap my head around why $C * k! \neq P$.

$C$ is the number of ways we can draw $k$ (not necessarily) items from $n$ distinct items with replacement. Here, order does not matter. To account for order mattering, I multiply by $k!$. So why isn't $C * k! = P$?

Best Answer

Because there are fewer orders if some of the elements match. If you have the combination $1,1,2$ there are only $3$, not $3!=6$ orders for these elements. You therefore cannot just multiply the number of combinations by $k!$ to get the number of permutations.

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