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Question: In how many ways can you sort 8 of 12 flags (4 flags for each country) so there will be at least one flag from each country? Final answer: 4620

I'm not sure whether it's possible, but I'd like to solve this question by reducing the impossible ways from the total options. I think that there are ${3 \choose 1}\dfrac{8!}{4! 4! 0! }$ impossible ways of sorting. How can the total options be calculated? I assume that the flags of the same country are indistinguishable.

Thanks!

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  • $\begingroup$ My bad, I'm in a bit of delay. Apologies and thanks for helping! $\endgroup$
    – Shahar
    Aug 8, 2012 at 7:53

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I don't see how to get the total options without adding up contributions from individual partitions of $8$. I'd do it like this:

$$ 3!\frac{8!}{4!3!1!}+3\frac{8!}{4!2!2!}+3\frac{8!}{3!3!2!}=4620\;. $$

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