# Combinatroics Problem Help

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, 5 months ago

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

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.

- 6 years, 5 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.

- 6 years, 5 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.

- 6 years, 5 months ago

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 $5!$ comes in, because order matters.

Staff - 6 years, 5 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.

- 6 years, 5 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!)

- 6 years, 5 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.

- 6 years, 5 months ago

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

- 6 years, 5 months ago

I think you are correct..

- 6 years, 5 months ago

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

- 6 years, 5 months ago