Consider $x$ containers each containing $n$ objects. A selection is the set realised by drawing $1$ item from each container.
How many unique selections can be made according to the specification?
My initial reasoning:
Let the containers be $C_i , i = 1, 2, \cdots, x \,$, there are $\binom{n}{1} = n$ ways to pick an item from $C_i$.
Let:
- $\mathbf{S}$ be the set of all selections,
- $\mathbf{s_i}$ denote a solution, and
- $\{a_j \in C_i : j = 1, \cdots, n\}$ be the $j^{th}$ element of $C_i$.
the number of selections in which $a_j$ appears is
\begin{equation} \begin{aligned} number\, of \, occurrences\ of \ a_j &= \begin{cases} n (x-1), &x>1 \\ x \end{cases}\\ & = \|\{a_j \in s_j \}: s_j \in \mathbf{s_i}\}\| \end{aligned} \end{equation}
Since $\|C_i\| = n$ and there are $x$ such sets, it follows that $\|\{a_j \in C_i : j = 1, \cdots, n\}\| = nx$. Multiplying the number of elements in $\{a_j \in C_i : j = 1, \cdots, n\}$ with the number of times each one occurs in the set of all solutions yields:
$$ \begin{cases} n^2 (x^2 - x), & x>1 \\ nx^2 \end{cases}\\ $$ But this cannot be the size of the set of all solutions because just multiplying (an empirical test with 3 containers and 2 objects show this is not $\|\mathbf{S}\|$).
I think since $\|C_i\| = n$, the correct approach is to multiply the number of occurrences of an object with the number of objects in one container. $$ \|\mathbf{S}\| = \begin{cases} n^2 (x - 1), & x>1 \\ nx \end{cases}\\ $$
My interpretation of this is as follows:
- for a single container, the number of possible selections are $\mathbf{nx}$
- for multiple containers, the number of possible selections are $\mathbf{n^2(x -1)}$