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I would like to ask your advice in the following problem:

We have 9 different books. With how many different ways are we able to put 4 red, 3 blue and 2 green labels in these books?

I tried to solve it as a combinations problem with 3 subgroups in my mind. I would appreciate your guidance in the specific problem. Thank you very much in advance.

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Hint: Consider a fixed permutation of labels on the $9$ books. Then consider how many ways you could get the same permutation by moving around the red labels. Then do the same for the blue labels and the green labels. Finally, if you can find out how many total permutations there are you will have your answer.

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  • $\begingroup$ Thank you very much. $\endgroup$ – Magda Oct 9 at 18:21
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Choose which of the books get a red label. From the remaining unlabeled books, choose which receive a blue label. The remaining books each get a green label.

This gives an answer of $\binom{9}{4}\binom{5}{3}\binom{2}{2}$. Equivalently phrased, this is equal to a multinomial coefficient $\binom{9}{4,3,2} = \frac{9!}{4!3!2!}$

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  • $\begingroup$ Thank you very much. $\endgroup$ – Magda Oct 9 at 18:21

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