Similar to Macmahon's theorem, why does $\sum t^{\text{maj}(w)}=\sum t^{\text{inv}(w)}$?

Question

In 1913 Percy MacMahon proved that the distribution of the major index on all permutations of a fixed length is the same as the distributin of inversions. I'm trying to understand the identity $$ \sum_{w\in S_n\cdot(1^{k_1},2^{k_2},\dots,r^{k_r})}x^{\text{maj}(w)}=\sum_{w\in S_n\cdot(1^{k_1},2^{k_2},\dots,r^{k_r})}x^{\text{inv}(w)} $$ where $k_1+\cdots+k_r=n$, and $w\in[r]^n$. With help, I've come to understand that if $w(n)=s$, then $w'=(1 2 \cdots r)^{r-s}\circ w$ has $\text{maj}(w')=\text{maj}(w)-(k_{s+1}+\cdots+k_r)$. I don't see how that leads to the result however. I've been searching around for MacMahon's theorem, but haven't found anything that actually proves this identity, and was hoping it could be resolved here. Thank you.

Answer 1

This paper by Dominique Foata gives a combinatorial proof of MacMahon’s result. It’s actually not horrendously complicated; take a look at it, and if you get stuck, feel free to ask for help.