# Counting in Two ways: $m^3 = 6 {m \choose 3} + 6 {m \choose 2} + m \quad \forall m \in \mathbb{N}$

I am having an issue in counting in two ways. How would I go about proving:

$$m^3 = 6 {m \choose 3} + 6 {m \choose 2} + m \quad \forall m \in \mathbb{N}$$

I can see that substituting formulas and simplifying would make the equality. I am not sure I can produce a combinatorial proof (but that is not required for the course I am in).

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Pick $3$ items from a set of $m$ elements, repetitions allowed, and keeping track of the order of the first, second, third item.

The left side clearly counts the number of ways to do this. The right side breaks it up by picking an unordered set and then ordering it. (There is one more thing going on with the coefficient of the ${m \choose 2}$ term --- I'll let you figure out what it is.)

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We use the fact that usual powers can be expressed as a combination of falling factorials $x^{(k)}$, as well as the relation $x^{(k)}=k!\binom{x}{k}$ to yield the series:

$$x^n=\sum_{j=0}^n \left\{n \atop j\right\}j!\binom{x}{j}$$

where $\left\{n \atop j\right\}$ is a Stirling subset number. (See page 262 of Concrete Mathematics for a justification.) For $n=3$, we then have

\begin{align*}x^3&=0!\left\{3 \atop 0\right\}\binom{x}{0}+1!\left\{3 \atop 1\right\}\binom{x}{1}+2!\left\{3 \atop 2\right\}\binom{x}{2}+3!\left\{3 \atop 3\right\}\binom{x}{3}\\&=1\times0\times\binom{x}{0}+1\times1\times\binom{x}{1}+2\times3\times\binom{x}{2}+6\times1\times\binom{x}{3}\\&=x+6\binom{x}{2}+6\binom{x}{3}\end{align*}

and presto!

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