How many ways are there to train the players?

Denote by $x_i$ $(1\leq i\leq20)$ the number of participants that trainer$_i$ is taking care of. Then there are nonnegative integers $y_i$ such that $$x_1=y_1,\quad x_2=x_1+1+y_2,\quad x_3=x_2+1+y_3,\quad \ldots\ ,$$ so that $$x_k=k-1+\sum_{j=1}^k y_j\qquad(1\leq k\leq20)\ .$$ Now we want $$210=\sum_{k=1}^{20} x_k=190 +\sum_{k=1}^n\sum_{j=1}^k y_j\ ,$$ which can be rewritten as $$\sum_{j=1}^{20}(21-j)y_j=20\ .\tag{1}$$ Let $z_l:=y_{21-l}$ $(1\leq l\leq20)$. Then $(1)$ amounts to $$\sum_{l=1}^{20} l\>z_l=20\ .\tag{2}$$ We need the number of solutions of $(2)$ in integers $z_l\geq0$. Each vector $(z_1,z_2,\ldots, z_{20})$ satisfying $(2)$ encodes a partition of $20$, whereby $z_l$ denotes the number of parts of size $l$. It follows that the number of such vectors is equal to the number of these partitions, which is $627$, according to Abramowitz & Stegun. Multiply this with $20!$ to assign the different trainers to the different workloads. But we have not yet taken care of the 210 different personalities that have to be trained. This would mean setting up a multinomial coefficient for each of the $627$ admissible workload schemes.