Please help me with this combinatorics problem -

**In how many ways can 22 distinct books be given to 5 students so that two students have 5 books each and another three students have 4 books each?**

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## Comments

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TopNewest\( {22 \choose 5,5,4,4,4} = \frac{22!}{5! \cdot 5! \cdot 4! \cdot 4! \cdot 4!} = 5 646 383 542 800 \)

It's a multinomial coefficient, i.e. the way of divide 22 different objects into 5 different groups of 5, 5, 4, 4 and 4.

Wikipedia: Multinomial Coefficient

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I think your solution is wrong.Since there are two 5's and three 4's we have to divide your solution by 2!*3! and also we have to multiply your answer by 5! because we can arrange the 5 different groups in 5! ways.

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I think you are correct..

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This is exactly the solution given in my textbook. But I don't agree with it because multiplying the solution by 5! makes no sense. It would mean that students who need 5 books are provided with 4 books and vice-versa.

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and

In one case, the order of the groups matter, and in the other case, the order doesn't. This is where multiplying by \(5!\) comes in, because order matters.

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Thanks... Got it. I was getting the same answer but was not sure.

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