Colouring juggling balls Please forgive the phrasing of this question, it was the only way I could think of that gets the problem across.  Any other re-phrasings would be appreciated!  The aim is to count how many different sets of juggling balls you can make with colored panels.  Each ball has $2$ colours, red and blue say, and there are four different colours in total.  You are trying to make $2n$ balls in total and you have $2$ panels of the first colour, $2n-2$ of the second, $2$ of the third and $2n-2$ of the fourth.  As an example let's consider the colours red, blue, green and yellow.  So you have 
$$
R,B,...,B,G,Y,...,Y 
$$ 
for the first panels of each ball, with $n-1$ blues and $n-1$ yellows.  We now take all permutations of this set, I believe there are 
$$
\frac{(2n)!}{(n-1)!(n-1)!},
$$
and pair them element-wise with the first set giving as an example for $n=4$
$$
\begin{align}
& R,B,B,B,G,Y,Y,Y \\
& R,B,Y,B,G,Y,B,Y
\end{align}
$$ 
so, to repeat, I would like to know how many unique sets of the $2n$ balls are possible.  And again if anyone knows of a simpler way to phrase the problem I'm open to suggestions!  Thanks.
 A: Since we are always using panels in pairs, we can just count pairs.  Let's say we have one pair each of red ($R$) and green ($G$), and $n-1$ pairs each of blue ($B$) and yellow ($Y$). 
Think about where the red and green panels can be assigned: Exactly one of the following sets of balls must be present in any solution:


*

*$\{RB,GB\}$

*$\{RY,GY\}$

*$\{RB,GY\}$

*$\{RY,GB\}$

*$\{RG\}$


In the first case, you are left with $n-3$ pairs of blue panels and $n-1$ pairs of yellow panels.  The second case is similar.  In the third and fourth cases, you are left with $n-2$ blue and $n-2$ yellow panel-pairs.  And in the final case, you are left with $n-1$ blue and $n-1$ yellow panel-pairs. 
Here's how to compute the number of possible sets of balls if you have $n-2$ blue panel-pairs and $n$ yellow panel-pairs.  I'll leave the rest to you. 
Note that once you determine the number of $YY$ balls in a set, everything else is determined.  The smallest number of $YY$ balls that we can have is 1 (not 0 - why?).  The largest number is $\lfloor (n-1)/2\rfloor$.  The number of possible arrangements containing the set of balls listed in (1) above is therefore $\lfloor (n-3)/2\rfloor$. 
The rest of the calculations are similar.
