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For example, say I have $3$ objects, $6$ slots, and each object must be chosen twice, how do I go about solving that?

Would it just be $\binom{6}{2} \binom{4}{2} \binom{2}{2}$ or am I thinking about it completely wrong?

Additionally, say we changed it so object $A$ shows up $3$ times, object $B$ $2$ times, and object $C$ $1$ time.

I don't really understand the intuition behind it, so an explanation of the thought process would be awesome.

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    $\begingroup$ You've got the right answer for the first part. Whatever reasoning you used should get you the answer for the second part as well. $\endgroup$ – Gerry Myerson Dec 4 '18 at 6:10
  • $\begingroup$ Say we wanted to simply choose at least 1 of each type, but no more than 3, how would I go about that? For the previous ones, I just choose 2 of each object from the remaining # of objects, but you can't really do that. I considered doing the star-bar method and just subtracting the possibilities that don't satisfy the condition, but I was unsure of how to generate those possibilities. $\endgroup$ – Flyrom Dec 4 '18 at 6:22
  • $\begingroup$ With those restrictions, it has to be 1, 2, 3 or else 2, 2, 2. You know how to handle the second case, and I bet you can handle the first. $\endgroup$ – Gerry Myerson Dec 4 '18 at 6:26
  • $\begingroup$ Making any progress? $\endgroup$ – Gerry Myerson Dec 5 '18 at 23:31
  • $\begingroup$ Yeah, since we know the 2,2,2 combination from above, 3 2 1 is similar I found. I just did $\binom{6}{3} \binom{3}{2} \binom{1}{1}$ + the 2,2,2 combination. It made sense in my head, let me know what you think! $\endgroup$ – Flyrom Dec 6 '18 at 3:41
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The way you go about thinking about this problem is that you have 6 positions at the start. Then you need to choose 2 of those 6 for the first type, then 2 of 4 for 2nd, and then 2 of 2 for third. So the end result is $\binom{6}{2} \binom{4}{2} \binom{2}{2}$

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