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18 persons are going to be divided into groups of 3. There are 4 persons that can not be in the same group. How many ways can it be done?

My attempts so far: Total number of combinations (if each person can be with any other person) = $$ \frac{C(18,3)*C(15,3)*C(12,3)*C(9,3)*C(6,3)*C(3,3)}{6!}=190590400$$ I have then tried to remove all combinatinos where 2 or more persons that are in the same group that should not.

I have also tried to firstly place these 4 persons in separate groups: $$ \frac{C(14,2)*C(12,2)*C(10,2)*C(8,2)*C(6,3)*C(3,3)}{6!}=210210 $$ That answer is totaly wrong, the right answer is 5045040

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  • $\begingroup$ If we first place the 4 persons in separate groups I think it becomes something like this: $C(17, 2) + C(15, 2) + C(13, 2) + C(11, 2) + C(9, 3) + C(6, 3) + C(3, 3)$ $\endgroup$ – A-Sharabiani Oct 18 '15 at 15:20
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Strong hint

Why don't you see what happens to your 2nd answer if you permute the "special" 4 between the groups they were placed in ....

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$ – BLAZE Oct 18 '15 at 17:44
  • $\begingroup$ I have given a very strong hint, practically spoon feeding. Since you haven't realized it, I'm writing it in the question also. $\endgroup$ – true blue anil Oct 18 '15 at 17:55
  • $\begingroup$ Okay, acknowledged $\endgroup$ – BLAZE Oct 18 '15 at 18:02

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