I'm sure this is well-known but I don't know where to look in order to find it.
Consider a permutation, e.g. $\sigma = 2 1 4 3$ in one-line notation. This corresponds to a monotone triangle via
\begin{array}{rcccccccc} \sigma(\{1,2,3,4\}) & = & 1 & \; & 2 & \; & 3 & \; & 4 \\ \sigma(\{1,2,3\}) & = & & 1 & & 2 & & 4 \\ \sigma(\{1,2\}) & = & & & 1 & & 2 \\ \sigma(1) & = & & & &2 \\ \end{array}
where each entry in the triangle is equal either to the number directly to its left or right above it. Of course, if any number is equal to the number above and to its left, then the same must be true of any number to its left (or else we'd have the same number twice in a row), so each row in the triangle (after the first, which is fixed) breaks into a left half and a right half:
\begin{array}{rcccccccc} \sigma(\{1,2,3,4\}) & = & 1 & \; & 2 & \; & 3 & \; & 4 \\ \sigma(\{1,2,3\}) & = & & 1 & & 2 & | & 4 \\ \sigma(\{1,2\}) & = & & & 1 & & 2 & | \\ \sigma(1) & = & & & | &2 & \\ \end{array}
Specifying the location of the break in each row clearly lets us reconstruct the triangle and thus the permutation. So we have e.g.
$$(2; 2; 0) \leftrightarrow 2143$$
meaning that the breaks in the second, third, and fourth rows come after the second, second, and first slots, respectively. Similarly, if $\tau = 3214$ then
\begin{array}{rcccccccc} \tau(\{1,2,3,4\}) & = & 1 & \; & 2 & \; & 3 & \; & 4 \\ \tau(\{1,2,3\}) & = & & 1 & & 2 & & 3 & | \\ \tau(\{1,2\}) & = & & | & 2 & & 3 & \\ \tau(1) & = & & & | & 3 & \\ \end{array}
so we have
$$(3; 0; 0) \leftrightarrow 3214$$
It's possible that the convention I'm using here is nonstandard by being off by one or inverted or something. The notation I'm using is almost certainly nonstandard. But I'm really just talking about the general idea.
This is very closely related to a Lehmer Code or an inversion table (slightly different than a Lehmer code but equivalent; I dont have a web reference, but you can look these up in Stanley's Enumerative Combinatorics. Alternatively, the wikipedia page for permutations has a slight aside on them I believe). Given $\pi=\pi_1\dots\pi_n$, we say a pair of indices $(i,j)$ is an inversion if $i<j$ and $\pi_i>\pi_j$. Fixing $j$, we let $d_j$ denote the number of inversions of the form $(i,j)$. Note that there are $j-1-d_j$ pairs of indices $(i,j)$ which are NOT inversions.
Now look at what happens when you move down from the top row. We are removing the element $\pi_n$, and the position of the break counts the number of indices $(i,n)$ with $\pi_i<\pi_n$ (why?). So the first element in your tuple corresponds to $n-d_n-1$. One can see how this process generalizes from one row to the next, so you are left with the tuple $$(n-d_n-1,(n-1)-d_{n-1}-1,\dots,2-d_2-1)$$ Note that we do not need a tuple of length $n$ because we know $d_1=0$. From this tuple you can easily reconstruct the Lehmer Code or the inversion table, from which you can reconstruct the permutation.