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
4 years, 7 months ago

No vote yet
3 votes

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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

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 - 4 years, 7 months ago

Log in to reply

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 - 4 years, 7 months ago

Log in to reply

I think you are correct..

Rushi Rokad - 4 years, 7 months ago

Log in to reply

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 - 4 years, 7 months ago

Log in to reply

@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 \(5!\) comes in, because order matters.

Calvin Lin Staff - 4 years, 7 months ago

Log in to reply

@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 - 4 years, 7 months ago

Log in to reply

@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 - 4 years, 7 months ago

Log in to reply

@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 - 4 years, 7 months ago

Log in to reply

@Pravas Patra 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.

Lokesh Sharma - 4 years, 7 months ago

Log in to reply

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

Lokesh Sharma - 4 years, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...