In how many ways can we distribute 20 black balls and 2 white balls (indistinguishable modulo color) in 5 numbered glasses such that the fifth one doesn't have more black balls than white balls?
My strategy was to break this problem into 6 incompatible scenarios according to how many balls we decide to put in the fifth glass: (i) no balls, (ii) 1W, (iii) 1W1B, (iv) 2W, (v) 2W1B, (vi) 2W2B.
Therefore, the are
$${23\choose20}{5\choose2} + {23\choose20}{4\choose1} + {22\choose19}{4\choose1} + {23\choose20} + {22\choose19} + {21\choose18} = 15 {23\choose20} + 5{22\choose19} + {21\choose18}$$
total distributions.
However, I was told each of these scenarios is equivalent to counting anagrams of $m,n,k$ copies of B, W, G respectively. But when we count anagrams of BBBBBBBBBBBBBBBBBBBBBBBBBWWGGG isn't the answer $\frac{25!}{20!2!3!}$? That's not the same as ${23\choose20}{5\choose2}$.
I'm seriously confused.
The partition into the six possible distinguished cases is correct.
We have that the number of ways to put $b$ black balls into $g$ labelled (numbered) glasses is equivalent to the number of weak compositions of $b$ into $g$ parts, which is $$ \binom{b+g-1}{b} $$
Similarly for the white balls.
Therefore $$ \eqalign{ & N = \sum\limits_{cases} {\left( \matrix{ b + g - 1 \cr b \cr} \right)\left( \matrix{ w + g - 1 \cr w \cr} \right)} = \cr & = \sum\limits_{0\, \le \,k\, \le \,\,j\, \le \,2} {\left( \matrix{ 20 - k + 3 \cr 20 - k \cr} \right)\left( \matrix{ 2 - j + 3 \cr 2 - j \cr} \right)} = \cr & = \sum\limits_{0\, \le \,k\, \le \,\,j\, \le \,2} {\left( \matrix{ 23 - k \cr 20 - k \cr} \right)\left( \matrix{ 5 - j \cr 2 - j \cr} \right)} = \cr & = \sum\limits_{0\, \le \,j\,,\,\,k\, \le \,2} {\left( \matrix{ 23 - k \cr 20 - k \cr} \right)\left( \matrix{ j - k \hfill \cr j - k \hfill \cr} \right)\left( \matrix{ 5 - j \cr 2 - j \cr} \right)} = \cr & = \sum\limits_{0\, \le \,k\, \le \,2} {\left( \matrix{ 23 - k \cr 20 - k \cr} \right)\left( \matrix{ 6 - k \cr 2 - k \cr} \right)} = \cr & = \sum\limits_{0\, \le \,k\, \le \,2} {\left( \matrix{ 23 - k \cr 3 \cr} \right)\left( \matrix{ 6 - k \cr 4 \cr} \right)} = \cr & = \left( \matrix{ 23 \cr 3 \cr} \right)\left( \matrix{ 6 \cr 4 \cr} \right) + \left( \matrix{ 22 \cr 3 \cr} \right)\left( \matrix{ 5 \cr 4 \cr} \right) + \left( \matrix{ 21 \cr 3 \cr} \right) = \cr & = 15\left( \matrix{ 23 \cr 3 \cr} \right) + 5\left( \matrix{ 22 \cr 3 \cr} \right) + \left( \matrix{ 21 \cr 3 \cr} \right) = \cr & = 15\left( \matrix{ 23 \cr 20 \cr} \right) + 5\left( \matrix{ 22 \cr 19 \cr} \right) + \left( \matrix{ 21 \cr 18 \cr} \right) = \cr & = 35595 \cr} $$ which fits with your computation.
Concerning the Anagrams analogy, it is correct that the case in which, e.g., the 5th glass is empty corresponds to the anagrams of $$[20 \times B, 5 \times W, 3 \times G]$$ since the $G$ stands for the separation between two consecutive of the four glasses.
And that is in fact equal to the weak compositions of $20$ into four parts, times the compositions of $5$ into four parts.
Now, instead, the multinomial $$ N_{0,0} = \left( \matrix{ 28 \cr 20,5,3 \cr} \right) = {{28!} \over {20!5!3!}} $$ which appears in the expansion of the trinomial $$ \eqalign{ & \left( {B + W + G} \right)^{\,28} = \cr & = \ldots + B^{\,20} W^{\,5} G^{\,3} + B^{\,19} WBW^{\,4} G^{\,3} + \ldots + \ldots \cr & \cdots + GB^{\,12} W^{\,2} B^{\,6} GW^{\,2} B^{\,2} WG + \cdots \cr & = \ldots + \left( \matrix{ 28 \cr 20,5,3 \cr} \right)B^{\,20} W^{\,5} G^{\,3} + \ldots \cr} $$ is going to count also how a certain number of white and black balls inside a single glass alternate in color, that is the situation in which the glass has a one-ball diameter and we distinguish not only by the content in black/white of each glass but also by the color layering.