How can I divide $2$ identical objects of one type, $2$ identical objects of second kind and $2$ identical objects of third kind into $3$ groups such that each groups contains only two objects.
The answer is $5$.
I tried the following:
Assume that I have objects A,A,B,B,C,C. I need to make three groups from these objects of size 2. Let us first do the calculation of the objects were distinct.
$6!/(2!×2!×2!×3!) $ which is equal to 15. Now some cases need to be eliminated. But I can't understand how can I eliminate them without actually making groups and ruling out identical cases.