Interpreting $n!$ as the volume of a $1 \times 2 \cdots \times n$ box

Question

Q. Are there relationships or proofs that are illuminated by viewing $n!$ as the volume of a $1 \times 2 \cdots \times n$ box in $n$-dimensions?

I cannot think of any, but perhaps they exist...?

Answer 1

Ok, so I'm writing this as an answer because it doesn't fit in the comments:

We can give a proof of $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$ using a volume interpretation but it's not that illuminating and it is not interesting enough (maybe). Even worse, I will have to rewrite the equation as $$n(n-1)\cdots (n-k+1)=k\cdot (n-1)(n-2)\cdots (n-k+1)+(n-1)(n-2)\cdots (n-k)$$

This can be interpreted as a volume summation if one shows that the $n\times (n-1)\times\cdots\times(n-k+1)$ box (call it $A$), has a tiling with $k$-many $1\times(n-1)\times\cdots\times(n-k+1)$ boxes (call them $d$) and one $(n-1)\times\cdots\times(n-k)$ box (call it $B$).

Indeed, say first the box $A$ is given by the set of vectors $$\big\{ (n-k+1,0,\cdots,0), (0,n-k+2,0,\cdots,),\cdots, (0,0,\cdots,0,n)\big\} $$ Now, place the box $B$ inside $A$ with gravity in the $(n-k+1,n-k+2,\cdots,n)$ direction (i.e. $B$ touches exactly half of the $2n$ sides of $A$ and shares exactly one vertex with it (the vertex $(n-k+1,n-k+2,\cdots,n)$. This would look somewhat like:

Now, notice that every point in the complement has a coordinate that belongs to the interval $[0,1]$. We can use that to give a stratification of the complement in $k$ boxes $B_1',B_2',\cdots, B_k'$, with respect to which coordinate is in $[0,1]$. Of course, some boxes overlap, that's why we define $B1=B1',\ B_2=B_2'-B_1', \cdots, B_i=B_i'-B_{i-1}'-\cdots -B_1'$ , etc.. to get an actual stratification.

Now, it's not very difficult to see that the volume of $B_i$ is exactly $(n-1)(n-2)\cdots (n-i+1)\cdot 1\cdot (n-i)\cdot (n-k+1)$, which is what we wanted. The formula for the volume is indicative of the proof, I feel it will get too long to read if I write it down...

Notice that this might help to answer my comment above, whether the box $n\times(n-1)\times\cdots\times(n-k+1)$ has a tiling with $\binom{n}{k}$-many $1\times 2\times\cdots\times k$ boxes. If we show that the complement we described can be tiled by $\binom{n-1}{k-1}$-many $1\times 2\times\cdots\times k$ boxes, we are done (by induction).

However, the particular stratification I chose doesn't work in this case as $(n-1)\times(n-2)\times\cdots\times (n-k+1)$ isn't always tilable. So, one should find a better stratification, or make an argument that might have to choose one depending on the particular $k$ and $n$...

Answer 2

Although not 100% about what you are asking (perhaps 87%), consider an $n$-dimensional cube with side $x$. The $n$-th derivative of the box's volume is $n!$. That is $$\frac{d^n}{dx^n} x^n=n!$$ Curiously enough, this is saying that the rate of the rate of the...of the rate of increase of the volume of an $n$-dimensional cube is precisely the volume of a $1\times2\times3\times\ldots\times n$ box. We can think about what's going on here in terms of a 3-dimensional cube to get a better handle. The first derivative of the volume with respect to the side length is just going to be three times the area of its faces. Why? Because if we increase the side length ($x$) by length $\delta$, we will have added three super-thin 'slices' to three of the faces of the cube. We can see this in the definition of the derivative: $$\frac{dx^3}{dx}=\lim_{\delta \to 0} \frac{(x+\delta)^3-x^3}{\delta}=\lim_{\delta \to 0}\frac{3x^2\delta+3x\delta^2+\delta^3}{\delta}$$ If we stop to think about what each of the terms in the numerator are describing, it is the following:

Therefore, the only surviving terms are those which describe the 3 cube faces. Now if we take the derivative of these three 'slices' (which are really just 2-dimensional squares in the limit as $\delta \to 0$), we see that the second derivative is telling us what the lengths of the two lines being added to three of the square sides of the cube are as we vary $x$ by an infinitesimal quantity. The third derivative is just telling us how many lines there are (actually it's the number of points we add to each line, but that's the same as the number of lines).

So putting it all together, we see that, in a sense, we can assign to each unit of volume in our box of dimensions $1,2,3,\ldots,n$ a point on an edge of an $n$-dimensional cube which would be added were the cube's side length to increase by $\delta$.

Answer 3

Consider the matrix $${\bf A_n} = \left(\begin{array}{cccc}1&0&\cdots&0\\0&2&\cdots&0\\\vdots&\vdots&\ddots&0\\0&0&0&n\end{array}\right)$$ It's determinant is $n!$. Furthermore

$$\det({\bf A_m}+k{\bf I_m}) = \frac{(m+k)!}{k!}$$ Which illuminates some relations to combinatorics (permutations).