Permutations with repetitions - which is k and which is n?

Question

I am learning permutations with repetitions, and working with the formula that $P(n,k)=n^k$. I understand the logic that with repetitions, we multiply $n$ by itself $k$ times. But in different problems, I find myself multiplying $k$ by itself $n$ times - getting the result $k^n$. So obviously I guess I'm not getting it.

Is there a simply rule to determine which set is the $n$ and which one is the $k$?

For example, the number of possibilities to divide $k$ objects into $n$ cells is $n^k$, because each object has $n$ choices. But how do I know not to turn it around, because my intuition here is actually that every cell has $k$ options, which is incorrect.

This is especially confusing because I had another exercise - "how many ways are there to put $0$ and $1$ in a row of $4$ places" - in this case, the "cells" were the $k$ and the "objects to choose from" were the $n$.

Thanks!

Answer 1

You can rule out the misguiding intuition by logic. For that intuition to work, it must have been the case, that the first cell can be filled in $(k+1)$ ways. (TRUE) For each such way, the second cell can also be filled in $(k+1)$ ways. Oops, that does not sound right any more does it? If the first cell has all the $k$ goods, how can the 2nd have $k$ goods too? So, you are only left with the right way to think about the problem.

For simplicity you can also use very basic examples to clear out the formula. Start by thinking of the easier counter part. "how many ways are there to put $0$ and $1$ in a row of $4$ places?" Now what is $n, k$ here? Start by thinking of the simpler question, "how many ways are there to put 0 and 1 in a row of 1 place?", the answer has to be $ 2=2^1\neq 1^2$. So, $n=2, k=1$ are identified through the simpler counterpart. For how many ways are there to put 0 and 1 in a row of 4 places, it has to be $n=2, k=4$.

Answer 2

Consider the following problem: License plates in Monkey Island consist of four digits. How many License plates are possible. So the license plates look like this:

$\underbrace x _{digit}$ $\underbrace x _{digit}$ $\underbrace x _{digit}$ $\underbrace x _{digit}$

So if we where to make ourselves our own license plate we could start deciding the first digit (10 options) then the second digit (another 10 options) and so on. At the end we would have made 4 decisions, and each time we made a decision the would have been 10 options. So there were $10\cdot10\cdot10\cdot10=10^4$ different "paths" we could have taken.

In this case we had $n=10$ and $k=4$. In general $n$ is the choices we have for each decision and $k$ is the number of decisions we have to take.

Answer 3

You have a number of distinct variable events which can each take on a number of values, independently.  A sequence of coins are flipped, or dice rolled, objects are drawn with repetition (coloured balls from a bag, cards from a deck, etc), distinct objects sorted into containers (darts thrown at a board, distinct balls into boxes, etc.) an so forth.

They are sometimes called trials and results.

Whether you call these $k$ and $n$ or call them $n$ and $k$, you count the number of variable events, and raise it to the power of the number of possible values each can have.

$\begin{align} P(\text{variables},\text{values}) & = (\text{variables})^{(\text{values})} \\ P(\text{trials},\text{results}) & = (\text{trials})^{(\text{results})} \\ P(n, k) & = n^k \\ P(8, 6) & = 8^6 & \text{eight dice are rolled} \\ P(42, 2) & = 42^2 & \text{forty two coins are flipped} \\ P(7, 52) & = 7^{52} & \text{seven cards are drawn from a deck with repetition} \\ P(100, 4) & = 100^4 & \text{one hundred darts are thrown at a board with four segments} \end{align}$

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For example, the number of possibilities to divide $k$ objects into $n$ cells is $n^k$ , because each object has $n$ choices. But how do I know not to turn it around, because my intuition here is actually that every cell has $k$ options, which is incorrect.

If distinct objects are "choosing" cells, then each ball faces a choice of one of every cell. We have distinct objects independently taking on a number of values.

If cells are "choosing" a number of objects, then each subsequent cell can choose only a number of objects from those not already claimed by another.