TITLE: Drawing balls from an urn or counting certain posets
QUESTION [9 upvotes]: A  colleague of mine was curious about the number of possible start-configurations in a game. The game itself is not known to me, but the question which he formulated as urn problem was interesting.

Urn problem:

Assume we have an urn containing 100 balls. The balls are colored, 25 are red, 25 blue, 25 green  and  25 are black. We pick four balls without replacement and repeat this step until the urn is empty.
We so obtain 25 groups of four balls each and the question is: how many configurations of this type are possible?   Thereby we may assume the order of the $4$ balls in each group is not relevant as well as the order of the $25$ groups is not relevant.

A reformulation:

Given an alphabet $V=\{1,2,3,4\}$ we consider $4$-letter words $x_1x_2x_3x_4$ with $1\leq       x_1\leq x_2\leq x_3\leq x_4\leq  4$ when considered as numbers.
These $4$-letter words are building blocks of words of length $100$. We take $25$ blocks of this kind to form a $100$-letter word $w=b_1b_2\ldots b_{25}$ with the property that $b_j\leq b_{j+1}, 1\leq j\leq 25$ when considered as numbers.

Question: How many words of this type contain exactly $25$ characters of each of the characters $j\in V, 1\leq j\leq 4$.
In general we are given an alphabet $V=\{1,2,\ldots,q\}$ with size $|V|=q$.

(a) We consider words of length $N$ and building blocks $x_1x_2\ldots x_M$ of size $M$ with $1\leq x_1\leq x_2\leq \cdots \leq x_M\leq  q$ and $M|N$, i.e. $N$ being an integer multiple of $N$.
(b) The words are of the form $w=b_1b_2\ldots b_{N/M}$ with $b_j\leq b_{j+1}, 1\leq j \leq N/M-1$.
(c) We are looking for the number  $\color{blue}{A_q(N,M)}$, the number of words as specified in (a) and (b) which contain $N/q$ characters of each of the characters $j\in V, 1\leq j\leq q$ implying that $N$ is an integer multiple of $q$ as well.

In this setting the urn problem is asking for $\color{blue}{A_4(100,4)}$.

The number of building blocks of $A_q(N,M)$ can be easily determined. It is
\begin{align*}
\sum_{1\leq x_1\leq x_2\leq\cdots\leq x_M\leq q}1=\binom{M+q-1}{M}\tag{1}
\end{align*}
A generating function of (1) can be easily derived. We have
\begin{align*}
\sum_{M=0}^\infty\sum_{q=0}^\infty   x^My^q\binom{M+q-1}{M}&=\frac{1-x}{1-x-y}\\
&=1+y\left(1+x+x^2+x^3+x^4+\cdots\right)\\
&\qquad+y^2\left(1+2x+3x^2+4x^3+5x^4+\cdots\right)\\
&\qquad+y^3\left(1+3x+6x^2+10x^3+\color{blue}{15}x^4+\cdots\right)\\
&\qquad\cdots
\end{align*}
Denoting with  $[x^M]$ the coefficient of $x^M$ in a series we see for instance $[x^4y^3]\frac{1-x}{1-x-y}=\binom{6}{2}=15$  which  is  the number of valid building  blocks of size  $4$  when given  a three letter alphabet $V=\{1,2,3\}$. These $15$ building blocks are
\begin{align*}
1111\quad1122\quad1222\quad1333\quad2233\\
1112\quad1123\quad1223\quad2222\quad2333\\
1113\quad1133\quad1233\quad2223\quad3333\\
\end{align*}
The difficult part  (at  least for me) is to determine the number of valid words $A_q(N,M)$ which  can  be generated from  these building  blocks. I've tried to derive a    generating   function  which  describes   this  scenario,  but I  wasn't   successful up to  now.

Posets: Another approach could  be  using posets based upon the approach:

Start  with  an  empty  word  and  append  $N/M$ times a  building  block respecting the ordering given in (b). 
Derive a generating function for the number of valid posets.

In order to better see the situation, here is a manageable example. We are looking for $A_2(12,M)$ the number of words of length $12$ from a two-letter alphabet with different block-sizes  $M$ following (a) - (c) from above. The Hasse-diagrams for $M=2,3,4,6$ are:

We see graded  posets  of length $N/M$ with  $A_2(12,2)=A_2(12,6)=4$   and  $A_2(12,3)=A_2(12,4)=5$  indicating the symmetry
  \begin{align*}
A_q(N,M)=A_q(N,N/M)
\end{align*}

Here is a  list  of small  values  of $A_2(N,M)$:
$$
\begin{array}{r|rrrrrr}
M&1&2\\
A_2(2,M)&1&1\\
\hline
M&1&2&4\\
A_2(4,M)&1&2&1\\
\hline
M&1&2&3&6\\
A_2(6,M)&1&2&2&1\\
\hline
M&1&2&4&8\\
A_2(8,M)&1&3&3&1\\
\hline
M&1&2&5&10\\
A_2(10,M)&1&3&3&1\\
\hline
M&1&2&3&4&6&12\\
A_2(12,M)&1&4&5&5&4&1\\
\end{array}
$$

Summary: The question  is how to find a formula for $A_q(N,M)$ or how to derive a generating function for these numbers. Alternatively is there an appropriate technique to count the number  of posets corresponding to $A_q(N,M)$?

REPLY [2 votes]: This answer is a  supplement to  the great answer  of  @MarkoRiedel and  a kind of reflection to better see  the   mechanisms working. The original problem was to determine $A_4(100,4)$ which can be written  according  to Markos  answer as
\begin{align*}
A_4(100,4)=[a^{25}b^{25}c^{25}d^{25}]Z\left(S_{25}(Z\left(S_4;a+b+c+d\right)\right)
\end{align*}
It is  a hopeless task to tackle this manually. But we can see all important aspects by taking a  smaller parameter  $N=8$ instead of $N=100$.  Leaving  the  other  parameter $M=4,q=4$  unchanged  we have  $N/M=N/q=2$ and we calculate
\begin{align*}
\color{blue}{A_2(8,4)=[a^{2}b^{2}c^{2}d^{2}]Z\left(S_{2}(Z\left(S_4;a+b+c+d\right)\right)}\tag{1}
\end{align*}

Calculation of  $Z(S_4)$:
We  start calculating $Z(S_4)$ by using the recurrence relation:
  \begin{align*}
Z(S_n)=\frac{1}{n}\sum_{j=1}^n a_j Z(S_{n-j})\qquad\textrm{where}\qquad Z(S_0)=1
\end{align*}
  We obtain
  \begin{align*}
Z(S_0)&=1\\
Z(S_1)&=\frac{1}{1}\sum_{j=1}^1a_jZ(S_{1-j})=a_1Z(S_0)\\
&=a_1\\
Z(S_2)&=\frac{1}{2}\sum_{j=1}^2 a_jZ(S_{2-j})=\frac{1}{2}\left(a_1Z(S_1)+a_2Z(S_0)\right)\\
&=\frac{1}{2}\left(a_1^2+a_2\right)\tag{2}\\
Z(S_3)&=\frac{1}{3}\sum_{j=1}^3 a_jZ(S_{3-j})=\frac{1}{3}\left(a_1\frac{1}{2}\left(a_1^2+a^2\right)+a_2a_1+a_3\right)\\
&=\frac{1}{6}\left(a_1^3+3a_1a_2+2a_3\right)\\
\color{blue}{Z(S_4)}&=\frac{1}{4}\sum_{j=1}^4a_jZ(S_{4-j})\\
&=\frac{1}{4}\left(a_1\frac{1}{6}\left(a_1^3+3a_1a_2+2a_3\right)+a_2\frac{1}{2}\left(a_1^2+a_2\right)+a_3a_1+a_4\right)\\
&\,\,\color{blue}{=\frac{1}{24}\left(a_1^4+6a_1^2a_2+8a_1a_3+3a_2^2+6a_4\right)}\tag{3}
\end{align*}

An ordinary generating function of the cycle index $Z(S(n))$ is 
\begin{align*}
\sum_{n=0}^\infty Z(S_n)) z^n=\exp\left(a_1z+\frac{a_2}{2}z^2+\frac{a_3}{3}z^3+\cdots\right)
\end{align*}
We can use this function to make a plausibility check of (3) via Wolfram Alpha by typing 

SeriesCoefficient[Exp[a_1*z+a_2*z^2/2+a_3*z^3/3+a_4*z^4/4],{z,0,4}]

which gives the expected result.

Calculation of $Z(S_4;a+b+c+d)$
We substitute $a+b+c+d$ in (2)
\begin{align*}
Z(S_4)=\frac{1}{24}\left(a_1^4+6a_1^2a_2+8a_1a_3+3a_2^2+6a_4\right)
\end{align*}
by replacing $a_j$ with $a^j+b^j+c^j+d^j$ and obtain
  \begin{align*}
\color{blue}{Z(S_4;}&\color{blue}{a+b+c+d)}\\
&=\frac{1}{24}\left((a+b+c+d)^4+6(a+b+c+d)^2(a^2+b^2+c^2+d^2)\right.\\
&\qquad\qquad \left.+8(a+b+c+d)(a^3+b^3+c^3+d^3)\right.\\
&\qquad\qquad \left.+3(a^2+b^2+c^2+d^2)^2+6(a^4+b^4+c^4+d^4)\right)\\
&=\cdots\\
&\,\,\color{blue}{=a^4+a^3b+a^3c+a^3d+a^2b^2+a^2bc+a^2bd+a^2c^2+a^2cd+a^2d^2}\\
&\qquad\color{blue}{+ab^3+ab^2c+ab^2d+abc^2+abcd+abd^2+ac^3+ac^2d+acd^2+ad^3}\\
&\qquad\color{blue}{+b^4+b^3c+b^3d+b^2c^2+b^2cd+b^2d^2+bc^3+bc^2d+bcd^2+bd^3}\\
&\qquad\color{blue}{+c^4+c^3d+c^2d^2+cd^3+d^4}\tag{4}
\end{align*}

Note that (4) has $35$ summands each with coefficient $1$, so that $Z(S_4;a+b+c+d)|_{a=b=c=d=1}=35$. This corresponds to (1) in the original post which is, since $M=4$ and $q=4$
\begin{align*}
\sum_{1\leq x_1\leq x_2\leq x_3 \leq x_4\leq 4}1=\binom{4+4-1}{4}=\binom{7}{3}=35
\end{align*}

Calculation of $[a^2b^2c^2d^2]Z(S_2;Z(S_4;a+b+c+d))$:
We substitute (4) in $Z(S_2)=\frac{1}{2}\left(a_1^2+a_2\right)$ according to (2) and we obtain
  \begin{align*}
\color{blue}{[a^2}&\color{blue}{b^2c^2d^2]Z(S_2;Z(S_4;a+b+c+d))}\\
&=[a^2b^2c^2d^2]\frac{1}{2}\left(a^4+a^3b+a^3c+a^3d+a^2b^2+a^2bc+a^2bd\right.\\
&\qquad\qquad\qquad\quad\left.+a^2c^2+a^2cd+a^2d^2+ab^3+ab^2c+ab^2d\right.\\
&\qquad\qquad\qquad\quad\left.+abc^2+abcd+abd^2+ac^3+ac^2d+acd^2\right.\\
&\qquad\qquad\qquad\quad\left.+ad^3+b^4+b^3c+b^3d+b^2c^2+b^2cd +b^2d^2 \right.\\
&\qquad\qquad\qquad\quad\left.+bc^3+bc^2d+bcd^2+bd^3+c^4+c^3d+c^2d^2\right.\\
&\qquad\qquad\qquad\quad\left.+cd^3+d^4\right)^2\\
&\quad+[a^2b^2c^2d^2]\frac{1}{2}\left(a^8+a^6b^2+a^6c^2+a^6d^2+a^4b^4+a^4b^2c^2+a^4b^2d^2\right.\\
&\quad\qquad\qquad\qquad\quad\left.+a^4c^4+a^4c^2d^2+a^4d^4 +a^2b^6+a^2b^4c^2+a^2b^4d^2\right.\\
&\quad\qquad\qquad\qquad\quad\left.+a^2b^2c^4+(abcd)^2+a^2b^2d^4+a^2c^6 +a^2c^4d^2+a^2c^2d^4 \right.\\
&\quad\qquad\qquad\qquad\quad\left.+a^2d^6+b^8+b^6c^2+b^6d^2+b^4c^4+b^4c^2d^2+b^4d^4\right.\\
&\quad\qquad\qquad\qquad\quad\left.+b^2c^6+b^2c^4d^2+b^2c^2d^4+b^2d^6
+c^8+c^6d^2+c^4d^4\right.\\
&\quad\qquad\qquad\qquad\quad\left.+c^2d^6+d^8\right)\\
&=[a^2b^2c^2d^2]\frac{1}{2}\left(a^2b^2+a^2bc+a^2bd+a^2c^2+a^2cd+a^2d^2\right.\\
&\qquad\qquad\qquad\quad\left.+ab^2c+ab^2d+abc^2+abcd+abd^2+ac^2d+acd^2\right.\\
&\qquad\qquad\qquad\quad\left.+b^2c^2+b^2cd+b^2d^2+bc^2d+bcd^2+c^2d^2\right)^2\\
&\quad+[a^2b^2c^2d^2]\frac{1}{2}(abcd)^2\tag{5}\\
&=[a^2b^2c^2d^2]\frac{1}{2}\left(2\left(a^2b^2\right)\left(c^2d^2\right)
+2\left(a^2bc\right)\left(bcd^2\right)
+2\left(a^2bd\right)\left(bc^2d\right)\right.\\
&\qquad\qquad\qquad\quad\left.+2\left(a^2c^2\right)\left(b^2d^2\right)
+2\left(a^2cd\right)\left(b^2cd\right)
+2\left(a^2d^2\right)\left(b^2c^2\right)\right.\\
&\qquad\qquad\qquad\quad\left.+2\left(ab^2c\right)\left(acd^2\right)
+2\left(ab^2d\right)\left(ac^2d\right)
+2\left(abc^2\right)\left(abd^2\right)\right.\\
&\qquad\qquad\qquad\quad\left.+(abcd)^2\right)\\
&\quad+[a^2b^2c^2d^2]\frac{1}{2}(abcd)^2\tag{6}\\
&\,\,\color{blue}{=10}\tag{7}
\end{align*}

Comment:

In (5) we keep only terms from the line above which have linear or square factors, since other terms do not contribute to $[a^2b^2c^2d^2]$.
In (6) we multiply out and indicate the factors by keeping them in parenthesis.


We finally got the result  (7):
  \begin{align*}
\color{blue}{A_4(8,4)}&=[a^2b^2c^2d^2]Z(S_2;Z(S_4;a+b+c+d))\\
&\,\,\color{blue}{=10}
\end{align*}
We can verify the result by listing the $35$ valid configurations according to $Z(S_4;a+b+c+d)$
$$
\begin{array}{cccc}
1111&1222&2222&3333\\
1112&\color{blue}{1223}&2223&3334\\
1113&\color{blue}{1224}&2224&\color{blue}{3344}\\
1114&\color{blue}{1233}&\color{blue}{2233}&3444\\
\color{blue}{1122}&\color{blue}{1234}&\color{blue}{2234}&4444\\
\color{blue}{1123}&\color{blue}{1244}&\color{blue}{2244}&\\
\color{blue}{1124}&1333&2333&\\
\color{blue}{1133}&\color{blue}{1334}&\color{blue}{2334}&\\
\color{blue}{1134}&\color{blue}{1344}&\color{blue}{2344}&\\
\color{blue}{1144}&1444&2444&\\
\end{array}
$$
The valid strings which have no more than two occurrences of each of the characters are marked $\mathrm{\color{blue}{blue}}$. Corresponding with
  $[a^2b^2c^2d^2]Z(S_2(Z(S_4;a+b+c+d))$ we concatenate the valid $\mathrm{\color{blue}{blue}}$ strings and find the $10$ resulting strings
  \begin{align*}
1122.3344\qquad1144.2233\\
1123.2344\qquad1223.1344\\
1124.2334\qquad1224.1334\\
1133.2244\qquad1233.1244\\
1134.2234\qquad1234.1234\\
\end{align*}