(-1)^{j+1}  q^{rtj+j}  \frac {  ( q^{-kt+t+1} - q^{-kt+1} )   }{ (1-q)^{n+1} }   \bigg)$$

$$= {\rm CT}_q \bigg(  \sum_{j=0}^n \sum_{i\geq0}  \binom nj \binom{n+i}{n}  (-1)^{j+1}  q^{rtj+j+i}   ( q^{-kt+t+1} - q^{-kt+1 } )  \bigg)$$

$$= \sum_{j=0}^n  \binom nj   (-1)^{j+1}  \left( \binom{n-rtj-j+kt-t-1}{n} - \binom{n-rtj-j+kt-1}{n}  \right).$$ As in the previous case, the formula is obtained with the convention that $\binom nk=0$ when $n<0$, but is true in general.

We also obtain a formula for the flag Eulerian numbers.

The flag Eulerian number is: $$A^{(r)}_{n,k} =         \sum_{j=0}^{ \lfloor (k-1)/r \rfloor } \binom nj  (-1)^{j+1}  (k-rj-1)^n  
      - \sum_{j=0}^{ \lfloor k/r \rfloor } \binom nj      (-1)^{j+1}  (k-rj)^n.$$

The number $ \text{Vol}(\mathcal{A}^{(r)}_{n,k}) = \frac{1}{n!} A^{(r)}_{n,k}$ can be obtained as the dominant coefficient of the Ehrhart polynomial. From the exact formula we have just obtained, this dominant coefficient is: $$\sum_{j=0}^{ \lfloor (k-1)/r \rfloor } \binom nj  \frac{ (-1)^{j+1} }{n!}  (k-rj-1)^n  
   - \sum_{j=0}^{ \lfloor     k/r \rfloor } \binom nj  \frac{ (-1)^{j+1} }{n!}  (k-rj)^n.$$ This is the announced formula up to the normalization factor $n!$. This could also be obtained from .

Note that the particular case $r=1$ gives a well-known formula: $$\begin{aligned}
  A_{n,k} &= \sum_{j=0}^k \binom nj (-1)^{j+1} ( (k-j)^n - (k-j+1)^n )  \\
          &= \sum_{j=1}^{k+1} \binom n{j-1} (-1)^{j} (k-j+1)^n - 