Intuition for volume of a simplex being 1/n!

Solution 1:

It's easier to see an alternate simplex has volume $1/n!$: The set of all points $(x_1,x_2,\dots,x_n)$ with $0\leq x_1\leq x_2\leq\cdots\leq x_n\leq 1$. That's because the volume measures the probability that a random sequence of $n$ real numbers is in sorted order, and (except with the probability zero case where some pair of values are equal) there are $n!$ ways to permute a set of $n$ values, and only one of them is sorted.

That this is the same volume as the original simplex is a little harder to see - these two simplices are not congruent, so their equal volume requires a little linear algebra.

There is a linear transformation between the two sending $0\leq x_1\leq x_2\leq \cdots x_n\leq 1$ to $(x_1,x_2-x_1,\dots,x_n-x_{n-1})$. The determinant of this linear transformation is $1$, so it preserves hyper-volumes.

Another approach is to ask how many ways can $n$ natural numbers $a_1,a_2,\dots,a_n$ be chosen so that $a_1+a_2+\dots a_n \leq m$. Standard combinatorics says this is $\binom{m+n}{n}$. And we see that $\frac{1}{m^n}\binom{n+m}{n}$ is an approximation of the hypervolume you want as $m$ gets large, by dividing the space into "hypercubes" of side length $\frac{1}{m}$, and this value approaches the volume as $m\to\infty$.

Solution 2:

Denote the volume of this simplex by $\sigma_n$. Foliating the simplex with "horizontal" hyperplanes $x_n=z$ $(0\leq z\leq 1)$ and applying Fubini's theorem we obtain $$\sigma_n=\int_0^1(1-z)^{n-1}\sigma_{n-1}\>dz={1\over n}\sigma_{n-1}\ .$$