How many ways are there to train the players? $210$ players participated the summer course.How many ways are there to choose a trainer out of $20$ trainers for each player so that every trainer have to train different number of people?
My attempt:If the trainers had to have at least one player to train then we would have at least $1+2+ \dots +20$ players which is equal to $210$ so the answer would be $\frac{210!}{20!19! \dots 1!}$ but the problem is that a trainer can have no players to train so we will have at least $190$ players and we have to check every case which is really hard.Any hints?
This question is in chapter "combination" of the book so I am looking for a proof using combination more than others.But others are acceptable too. 
 A: 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.
A: We can solve the basic question - where every trainer has at least one person to train - in two steps. Imagine a bag of $210$ balls of different colours, with a different number of each colour. The players each take a ball. Now the different ways that can happen are given by the appropriate multinomial coefficient
$$\frac{210!}{20!\cdot 19!\cdot 18!\cdots 3!\cdot 2!\cdot 1!} = \frac{210!}{\prod_{k=1}^{20}k!}$$
Then we can allocate the colours to trainers in a second drawing, which can happen in $20!$ ways. This is multiplied in to give $$ \frac{210!\cdot 20!}{\prod_{k=1}^{20}k!}= \frac{210!}{\prod_{k=1}^{19}k!}$$
There are many ways to partition 210 into twenty different numbers if we allow a trainer to have zero trainees. We can consult a(20) for OEIS A000041 to get $627$ options - one of which we have already covered above. [This is the count of different multisets of integers that sum to $20$ - which we would then add on to a "preload" of $19,18,\ldots,1,0$ in the different partitions to get the distinct sums required.] Each of these would require similar treatment to above.
