Combinatroics Problem Help

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?

Note by Lokesh Sharma
6 years, 1 month ago

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3 votes

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(225,5,4,4,4)=22!5!5!4!4!4!=5646383542800 {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 - 6 years, 1 month ago

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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.

shiva raj - 6 years, 1 month ago

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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.

Lokesh Sharma - 6 years, 1 month ago

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@Lokesh Sharma You are forgetting that the students are distinct. There is a difference between:

How many ways are there to split 4 books into a group of 3 and a group of 1?

and

How many ways are there to distribute 4 books to 2 students, such that one student receives 3 books and the other student receives 1 book?

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

Calvin Lin Staff - 6 years, 1 month ago

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@Calvin Lin 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.

Lokesh Sharma - 6 years, 1 month ago

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@Calvin Lin 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!)

Pravas Patra - 6 years, 1 month ago

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@Pravas Patra No, it would be 22!5!5!4!4!4!2!3! \frac {22!}{5! 5! 4! 4! 4! 2! 3!} in case of dividing into groups and 22!5!5!4!4!4!2!3!5! \frac {22!}{5! 5! 4! 4! 4! 2! 3!} * 5! in above asked student case.

Lokesh Sharma - 6 years, 1 month ago

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@Calvin Lin Oh yes, I thought that it didn't matter which student has 5 books and which 4! Thanks for the correction btw ;)

Luca Bernardelli - 6 years, 1 month ago

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I think you are correct..

Rushi Rokad - 6 years, 1 month ago

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

Lokesh Sharma - 6 years, 1 month ago

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