Multinomial theorem with imposed conditions 
The number of ways in which 12 identical balls can be grouped in four marked non-empty sets $P, Q, R, S$ such that $n(P) < n(Q)$ is?

The answer to the above problem is the number of positive integral solutions of the expression $$P + Q + R + S = 12$$
when $P<Q$. I know that with no conditions imposed, the number of solutions is $^{12-1}C_{4-1} = ^{11}C_3$. How would one account for the condition when $P<Q$? Please explain the concept in detail in addition to providing a solution to the problem (which is given just as an example).
 A: So first of all, the solution without that condition is $\binom{12-1}{4-1}$. Since the sets are all non - empty.
Notice that $Q-P$ is a positive integer as well. The number of solutions with condition imposed is the number of positive integer solutions of
$2P+(Q-P)+R+S=12$
Since the number is small, You can count solutions for $2P=2,4,6,8$.
$2P=2,\ \ (Q-P)+R+S=10$ there are $\binom{10-1}{3-1}=36$ solutions
$2P=4,\ \ (Q-P)+R+S=8$ there are $\binom{8-1}{3-1}=21$ solutions
$2P=6,\ \ (Q-P)+R+S=6$ there are $\binom{6-1}{3-1}=10$ solutions
$2P=8,\ \ (Q-P)+R+S=4$ there are $\binom{4-1}{3-1}=3$ solutions
Total $70$ solutions
A: Let $a,b,c,d\ge0$. Then set $|P|=a+1$, $|Q|=a+b+2$, $|R|=c+1$, $|S|=d+1$ where $2a+b+c+d+5=12$. We want to count the non-negative solutions to
$$
2a+b+c+d=7\tag1
$$
Considering the coefficients of $x^n$ in
$$
\overbrace{\left(1+x^2+x^4+x^6+\dots\right)\vphantom{{x^2}^3}}^a\overbrace{\left(1+x+x^2+x^3+\dots\right)^3}^{b,c,d}\tag2
$$
$(1)$ gives a generating function of
$$
\begin{align}
&\frac1{1-x^2}\left(\frac1{1-x}\right)^3\\
&=\frac1{1+x}\left(\frac1{1-x}\right)^4\\
&=\frac12\left(\frac1{1-x}+\frac1{1+x}\right)\left(\frac1{1-x}\right)^3\\
&=\frac12\left(\frac1{1-x}\right)^4+\frac14\left(\frac1{1-x}+\frac1{1+x}\right)\left(\frac1{1-x}\right)^2\\
&=\frac12\left(\frac1{1-x}\right)^4+\frac14\left(\frac1{1-x}\right)^3+\frac18\left(\frac1{1-x}+\frac1{1+x}\right)\frac1{1-x}\\
&=\frac12\left(\frac1{1-x}\right)^4+\frac14\left(\frac1{1-x}\right)^3+\frac18\left(\frac1{1-x}\right)^2+\frac1{16}\left(\frac1{1-x}+\frac1{1+x}\right)\tag3
\end{align}
$$
Therefore,
$$
\begin{align}
&\left[x^n\right]\frac1{1-x^2}\left(\frac1{1-x}\right)^3\\[3pt]
&=\textstyle\frac12(-1)^n\binom{-4}{n}+\frac14(-1)^n\binom{-3}{n}+\frac18(-1)^n\binom{-2}{n}+\frac1{16}(-1)^n\binom{-1}{n}+\frac1{16}\binom{-1}{n}\\[3pt]
&=\frac12\binom{n+3}{n}+\frac14\binom{n+2}{n}+\frac18\binom{n+1}{n}+\frac1{16}+\frac1{16}(-1)^n\\
&=\frac{4(n+3)(n+2)(n+1)+6(n+2)(n+1)+6(n+1)+3+3(-1)^n}{48}\\
&=\frac{(4n+14)(n+3)(n+1)+3\left(1+(-1)^n\right)}{48}\tag4
\end{align}
$$
For $n=7$, we get
$$
\frac{42\cdot10\cdot8}{48}=70\tag5
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\sum_{P = 1}^{\infty}\sum_{Q = 1}^{\infty}
\sum_{R = 1}^{\infty}\sum_{S = 1}^{\infty}\bracks{P < Q}
\bracks{z^{12}}z^{P + Q + R + S}}
\\[5mm] = &\
\bracks{z^{8}}\sum_{P = 0}^{\infty}z^{P}
\sum_{Q = 0}^{\infty}z^{Q}\bracks{Q > P}
\sum_{R = 0}^{\infty}z^{R}\sum_{S = 0}^{\infty}z^{S}
\\[5mm] = &\
\bracks{z^{8}}\pars{1 - z}^{-2}\sum_{P = 0}^{\infty}z^{P}
\sum_{Q = P + 1}^{\infty}z^{Q} =
\bracks{z^{8}}\pars{1 - z}^{-2}\sum_{P = 0}^{\infty}z^{P}\,
{z^{P + 1} \over 1 - z}
\\[5mm] = &\
\bracks{z^{7}}\pars{1 - z}^{-3}\sum_{P = 0}^{\infty}z^{2P} =
\bracks{z^{7}}\pars{1 - z}^{-3}\,{1 \over 1 - z^{2}}
\\[5mm] = &\
\bracks{z^{7}}\pars{1 - z}^{-4}\pars{1 + z}^{-1} =
\sum_{k = 0}^{7}\bracks{z^{7}}\pars{1 - z}^{-4}\pars{-1}^{k}z^{k}
\\[5mm] = &\
\sum_{k = 0}^{7}\pars{-1}^{k}\bracks{z^{7 - k}}\pars{1 - z}^{-4} =
-\sum_{k = 0}^{7}\pars{-1}^{k}\bracks{z^{k}}\pars{1 - z}^{-4}
\\[5mm] = &\
-\sum_{k = 0}^{7}\pars{-1}^{k}{-4 \choose k}\pars{-1}^{k} =
-\sum_{k = 0}^{7}\pars{-1}^{k}{k + 3 \choose 3}
= \bbx{\Large 70} \\ &
\end{align}
