Distributing two distinct objects to identical boxes Hiii,
I've been struck with a problem which deals with the distribution of two distinct objects such that p of one type and q of other type into three identical boxes.
As if it were only one object with q copies i'd have used integer partitioning, and if all objects were distinct then could use Stirling Number of second kind. But what about this mixed case...?
Thnx!

For the sake of completeness I am posting an answer using the Polya Enumeration Theorem. As was pointed out, the group acting on the boxes is $S_3$ with cycle index
$$Z(S_3) = \frac{1}{6} a_1^3 + \frac{1}{2} a_1 a_2 + \frac{1}{3} a_3.$$
The pattern repertoire is given by
$$ \frac{1}{1-X}\frac{1}{1-Y}.$$
Substituting into the cycle index it now follows that the desired value is
$$ [X^p Y^q]  \left(
\frac{1}{6} \frac{1}{(1-X)^3}\frac{1}{(1-Y)^3} + 
\frac{1}{2} \frac{1}{1-X}\frac{1}{1-Y}  \frac{1}{1-X^2}\frac{1}{1-Y^2} +
 \frac{1}{3} \frac{1}{1-X^3}\frac{1}{1-Y^3} \right).$$
Now the first term (see the formula at the end) contributes
$$ \frac{1}{6} {p+2 \choose 2} {q+2 \choose 2}.$$
The second term combines the initial segment of the series for $1/(1-X^2)$ of degree at most $p$ with a term from $1/(1-X)$ and the initial segment of the series for $1/(1-Y^2)$ of degree at most $q$ with a term from $1/(1-Y)$ and hence contributes
$$ \frac{1}{2} 
\left(\sum_{m=0}^{\lfloor p/2\rfloor} 1 \right)
\left(\sum_{m=0}^{\lfloor q/2\rfloor} 1 \right) =
 \frac{1}{2} \lfloor (p+2)/2\rfloor \lfloor (q+2)/2\rfloor$$
and the third term contributes $1/3$ if $p$ and $q$ are both divisible by three because $1/(1-X^3)$ only contains terms of degree divisible by three and $1/(1-Y^3)$  also only contains terms of degree divisible by three.
This confirms the first answer that was posted, namely
$$\frac{1}{6} {p+2 \choose 2} {q+2 \choose 2}
+  \frac{1}{2} \lfloor (p+2)/2\rfloor \lfloor (q+2)/2\rfloor
+ \frac{1}{3} [[p\equiv 0 (3); q\equiv 0 (3)]].$$
Here we have made use several times of the fact that
$$[z^k] \frac{1}{(1-z)^m} = {k+m-1\choose m-1}.$$
