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

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...?

• Can we at least show that there is a tiling of a $(k+1)\times (k+2)\cdots \times (k+n)$ box with $1\times 2\times \cdots \times n$ boxes and in this way illustrate the fact that $\binom{k+n}{n}$ is an integer? Commented Apr 7, 2015 at 5:34
• @Theo I'm not sure whether that's even true; I can't quite visualize a tiling of a 5x6x7 box with 1x2x3 boxes (of course, as is often the case in math, if I had enough cubes and enough tape, perhaps I could be convinced otherwise) Commented Apr 11, 2015 at 19:13
• Five 1x2x3 dominoes make a 5x2x3 stack. Three of those makes 5x6x3, and two of those makes 5x6x6. That leaves 5x6x1 to be filled, and that's a 2-dimensional problem: tile a 5x6 rectangle with 2x3 rectangles.
– user21467
Commented Apr 11, 2015 at 19:24
• @Meelo I think it works for small values of $n$ but I cannot prove the general case... It is of course not true that if $ab$ divides $mn$, there are always suitable tilings (e.g. $3\times 8$ and $4\times 12$) but I think it should work for the factorial boxes... Commented Apr 11, 2015 at 20:15
• This is maybe only a little related, but the proof I know that there exists a solution to a 1st order, linear, vector-valued ODE of the form $X'(t) = A(t)X(t)$ uses the fact that that the simplex $\{ (x_1,\ldots,x_n) : 0 \leq x_1 \leq \ldots \leq x_n \leq 1 \}$ has $n$-dimensional volume $\frac{1}{n!}$. Commented Apr 12, 2015 at 1:01

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 $n$...

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 $\delta$.

• This is exactly the geometric representation of calculus by one historical Italian mathematician that i read on wikipedia. unfortunately i don't remember his name Commented Apr 18, 2015 at 0:53
• Let me know if you remember =) Commented Apr 18, 2015 at 3:00
• Was it Cavalieri? Commented Apr 18, 2015 at 3:54
• @BennettGardiner If you want that another user receives a notification of your comment, you should use @username syntax. (With some exception, when user is notified anyway. If you follow the above link, you can find more details - including explanation which users can be notified in this way.) Commented Jul 31, 2016 at 7:23
• Yes. Sorry for long delay, it must be him. Here i found wikipedia site en.wikipedia.org/wiki/Cavalieri%27s_quadrature_formula Commented Jul 31, 2016 at 19:09

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).