Can the $n$ in the binomial coefficient represent something different than the $n$ choices in $n^r$ outcomes? In a setup with $6$ customers distributed among three floors of a restaurant, the probability of having exactly one customer dine on the 1st floor is $(6c1)(1/3)^1(2/3)^5$.
Using the counting method, I imagine that using the formula $n^r$ for number of different outcomes means that $n$ (the number of choices) represents floors and that $r$ (the number of stages or trials) represents people. However, I have only seen the $n^r$ formula in the context of permutations (without repetition), whereas this does not seem to be a problem of permutations (as multiple customers can take on the same floor). While I understand that the number of different outcomes is $3^6$ because $n_1 \cdot n_2 \cdot n_3 \cdot n_4 \cdot n_5 \cdot n_6$ represents $n=3$ choices at each of $r=6$ stages, is it incorrect to think of this as $n^r$ different outcomes?
As far as $n$ taking on a different meaning in the binomial coefficient, there are $(6c1)$ ways to have $1$ customer dine on a particular floor. I take this to mean that due to the binomial coefficient being represented by $(nck)$, that $n$ in this context represents elements (rather than $n$ choices as above using the counting method), of which only $k=1$ are picked to take on a value (floor).
Is this an accurate interpretation of the representation of $n, r,$ and $k$ in the counting method in the binomial coefficient? I feel like trying to think of $n$ as the same entity in both formulas $(n^r)$ and $(nck)$ caused confusion for me between what a stage/trial is and what an element is in this context.
 A: This is a binomial distribution problem.  The probability of exactly $k$ successes in $n$ trials is
$$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$
where $p^k$ is the probability of exactly $k$ successes, $(1 - p)^{n - k}$ is the probability of exactly $n - k$ failures, and $\binom{n}{k}$ represents the number of ways exactly $k$ successes could occur in $n$ trials.
In this problem, a "success" is defined as a customer choosing to eat on the first floor.  Under the assumption that a customer is equally likely to choose any of the floors, $p = 1/3$ since there are three floors and the customer is choosing one of them.  Thus, $1 - p = 2/3$.
You are correct that $n$ represents the number of trials.  In this problem, each trial represents the choice a customer makes to eat on the first, second, or third floors.  The number $k$ represents the number of customers who select the first floor.  The factor $1/3$ is the probability that a customer chooses to eat on the first floor.  The exponent $k = 1$ represents the number of customers who opt to eat on the first floor. The factor $2/3$ is the probability that a customer does not eat on the first floor.  The exponent $5 = 6 - 1 = n - k$ is the number of customers who opt to not eat on the first floor.
You are correct that there are $3^6$ options in the sample space since each of the six customers has three choices.  The number of favorable cases is $6 \cdot 2^5$ since there are six customers who could choose to eat on the first floor and each of the five customers who choose not to eat on the first floor have two choices of floor on which to eat.
