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?**

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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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 – Luca Bernardelli · 3 years, 11 months ago

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– Shiva Raj · 3 years, 11 months ago

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.Log in to reply

– Rushi Rokad · 3 years, 11 months ago

I think you are correct..Log in to reply

– Lokesh Sharma · 3 years, 11 months ago

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.Log in to reply

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. – Calvin Lin Staff · 3 years, 11 months ago

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– Lokesh Sharma · 3 years, 11 months ago

I get it now. Thanks a lot, Calvin. I might have spent my whole life believing in that. Thank you very much for correcting me Sir.Log in to reply

– Luca Bernardelli · 3 years, 11 months ago

Oh yes, I thought that it didn't matter which student has 5 books and which 4! Thanks for the correction btw ;)Log in to reply

– Pravas Patra · 3 years, 11 months ago

u mean to say that the answer is 22! / {(5!⋅5!⋅4!⋅4!⋅4!)* 5!} due to 5 students ,,,,,,,,, and if in the case of 5 groups or 5 sets it will be 22!/(5!⋅5!⋅4!⋅4!⋅4!)Log in to reply

– Lokesh Sharma · 3 years, 11 months ago

No, it would be \( \frac {22!}{5! 5! 4! 4! 4! 2! 3!} \) in case of dividing into groups and \( \frac {22!}{5! 5! 4! 4! 4! 2! 3!} * 5!\) in above asked student case.Log in to reply

– Lokesh Sharma · 3 years, 11 months ago

Thanks... Got it. I was getting the same answer but was not sure.Log in to reply