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.

  • $\begingroup$ Have you tried simply enumerating the possibilities? $\endgroup$ – lulu Jan 30 '16 at 15:22
  • $\begingroup$ Yes. But I need the logic so that I can solve it if the number of objects or groups is changed. $\endgroup$ – devang singhi Jan 30 '16 at 15:25
  • $\begingroup$ the way the site works, you'll get a much better response from people if you write out your thoughts and indicate where you're stuck. $\endgroup$ – lulu Jan 30 '16 at 15:27
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    $\begingroup$ Hint: try to work recursively. Either your arrangement has duplicates (groups like $AA$) or it doesn't. If it does, then you are down to fewer types of objects. If it doesn't, then you have a "derangement", en.wikipedia.org/wiki/Derangement $\endgroup$ – lulu Jan 30 '16 at 15:34
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    $\begingroup$ Once again, I suggest editing your post to show your thoughts and efforts. It is difficult to help when we have no idea what level you are at. $\endgroup$ – lulu Jan 30 '16 at 15:42

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