where is the internal order in each cell is inside the formula $k^n$?

Question

$n$ different balls, $k$ different cells. In how many ways I can put the balls in the cells while the order in each cell is important?

Proposed solution: $k^n$.

My question - In the above formula, where is the internal order in each cell is calculated?

For example, if I arrange balls: $\{B_1,B_2,B_3\}$ in bins $\{C_1,C_2,C_3,C_4\}$ from the first ball $B_1$ to the last $B_3$, then there will be never any event which contain a cell where $B_1$ is not the first one inside the internal order.

Is this makes any sense? Or I don't understand the definition of internal order inside a cell?

Answer 1

With the help of JMoravitz, the currect answer is (using stars and bars, and then multiply the answer by the amopunt of starts permutations):

$$ \binom{n+k-1}{n}n! = \frac{(n+k-1)!}{(k-1)!} $$

Answer 2

As shown above, the solution is the rising factorial $k^{(n)}$

Let's say we have $n$ books and $(k-1)$ separators, such that we will form $k$ shelves (some of them could be empty).

There are $(n+k-1)!$ possible orders of the $(n+k-1)$ items. But the order of the $(k-1)$ separators does not matter, hence the answer :

$k^{(n)} = k.(k+1).(k+2)\dots (n+k-1) = \frac{(n+k-1)!}{(k-1)!}$