Falling factorial counts permutations, what does rising factorial count? Rising factorial example: Let $x = 7$ and $r = 4$. Then $7^{(4)} = 7(8)(9)(10) = 5040$. If we divide $7^{(4)}$ by $4!$ it counts multisubsets.
But what kind of combinatorial problem does rising factorial solve on its own?
 A: Imagine constructing a permutation of $k$ elements with $n-1$ elements initially, but add a spot for each number placed in the sequence instead of taking one away. I'm going to call this a rising permutation for lack of a better name.

You can visualize the construction of an ordinary permutation on $n$ elements by imagining $n$ slots and replacing one slot with a number every turn. We'll number the turns $1,2, \dots, n$ . In this array below each $*$ is an as-yet-unfilled box and each row represents the state at the end of the $n$th turn, with the first row being the initial state.
$$
\begin{array}{ccc}
* & * & * \\
* & 1 & * \\
2 & 1 & * \\
2 & 1 & 3 \\
\end{array}
$$
There are several equivalent ways of thinking about constructing a permutation one element at a time. This way doesn't make it easy to visualize what happens if there are more numbers than slots (the falling factorial case). However, it does make the analogy with a "rising permutation" more obvious. The key difference is that, on the $i$th turn, we place the number $i$ between one of the items already in the sequence. Doing so creates a slot instead of consuming one.
Each row represents a turn. The gaps in the earlier rows are so that each column tracks the lifetime of a single item in the sequence.
$$ \begin{array}{cccccc}
  & * & * &   & * &   \\
  & * & * & 1 & * &   \\
2 & * & * & 1 & * &   \\
2 & * & * & 1 & * & 3 \\
\end{array}$$
The above example is from $(4)^{\overline{3}}$ .
The rising factorial $(n)^{\overline{k}}$ asks how many distinct sequences can be made after $k$ turns if there are $(n-1)$ stars initially (equivalently $n$ stars at the end of turn $1$) . Note that a sequence contains the stars that are in it so * 1 * and 1 * * are not the same sequence.
A: This answer is just a rephrasing of Gregory Nisbet's earlier answer, but I hope that some people may find it clearer.
The rising factorial $(x)^k:= x(x+1)\dotsb (x+k-1)$ is the number of ways of placing $k$ flags (numbered $1,\dotsc,k$) on $x$ flagpoles.
for example
|3 |  |5 |1
|2 |  |  |4
|  |  |  |
|  |  |  |

is a configuration of five flags on four flagpoles. The order in which the flags appear on each flagpoles matters.
For the first flag there are $x$ positions (just choose any one of $x$ flagpoles). For the second flag, you have to choose which pole, and then if that pole already has a flag on it, then you have to choose whether to put the flag above or below that flag.
In general, if $r$ flags have already been placed, then there are $x+r$ choices for the $(r+1)$st flag--either at the top of one of $x$ poles, or just beneath one of the $r$ flags already placed (on the same pole).
It follows that $(x)^k$ is the number of ways of placing $k$ flags (numbered $1$ to $k$ on $x$ flagpoles).
