Combinatorial reasoning for linear binomial identity I have the following equation:
\begin{equation}
m^4 = Z{m\choose 4}+Y{m\choose 3}+X{m\choose 2}+W{m\choose 1}
\end{equation}
I iteratively took $m=1$ to $m=4$ to solve for the coefficients. I got the following values:
$Z=24$, $Y=36$, $X=14$ and $W=1$.
I then checked the equation with $m=5$ and "verified" the identity. 
However, what is the interpretation of these values? I am looking for a combinatorial argument of this equation. What do the values of W, X, Y, and Z mean here? I do not see any similarities/patterns between W, X, Y, and Z.


*

*What is the combinatorially interpretation of these values

*Is there a better way to look at this problem or enumerate?

*What else can I try?


All help is greatly appreciated!
 A: These are the numbers of ways in which you can form $4$-tuples of $m$ different items with $4$, $3$, $2$ and $1$ different entries, respectively. With $1$ entry, you only have $1$ choice, so $W=1$. With $2$ different entries, you can either have $1$ of one and $3$ of the other ($4$ possibilities), $2$ of both ($6$ possibilities) or $3$ of one and $1$ of the other (another $4$ possibilities), for a total of $X=14$ possibilities. With $3$ different entries, you necessarily have $2$ of one and $1$ of each of the other two; you have $3$ choices for which one to have $2$ of, and then $12$ ways to arrange them, so $Y=3\cdot12=36$; and with $4$ different entries you have $4!=24$ ways to arrange them. There are $m^4$ $4$-tuples in total, and $\binom mk$ ways to choose $k$ different items.
Also, note that the Stirling numbers of the second kind satisfy the identity
$$
\sum_{k=0}^n\left\{n\atop k\right\}(x)_k=x^n\;,
$$
where $(x)_k$ is the Pochhammer symbol,
$$(x)_k=x(x-1)(x-2)\cdots(x-k+1)=k!\binom xk\;,$$
so we have 
$$
\sum_{k=0}^nk!\left\{n\atop k\right\}\binom xk=x^n\;,
$$
so your coefficients are given by $\displaystyle k!\left\{4\atop k\right\}$ for $k=1,\dotsc,4$.
These two viewpoints are related in that the Stirling numbers of the second kind count the number of partitions of $n$ items into $k$ sets, and for each such partition you can choose $k$ out of the $m$ items and assign them to the sets of the partition in $k!$ different ways.
