Question 1145431: A library has 4 books from each of 3 different authors.
How many ways can the library arrange the books on the shelf if books by the same author must be placed together?
Answer by ikleyn(52806)   (Show Source):
You can put this solution on YOUR website! .
The problem's formulation is incomplete.
From the given formulation, it is NOT CLEAR if the books by the same author are distinguishable or not
(inside each group of 4 books).
I just asked this question in my previous post, but didn't get an answer.
It looks like, on the other side they do not understand the meaning of my question,
and posted the problem again with no change.
OK. I will give two solutions, for two extreme cases.
Solution 1. The books written by the same author (4 books in each group) are assumed as INDISTINGUISHABLE
In this case, we can consider each group as one aggregated block.
Then we can permute 3 such blocks in 3! = 3*2*1 = 6 ways.
The permutations inside each block do not produce nothing new, since the books of the same author are assumed indistinguishable.
Thus the answer in this case is 6 distinguishable arrangements.
This case is completed.
Solution 2. All the books written by the same author (4 books in each group) are assumed as DISTINGUISHABLE
In this case, again, we can consider each group as one aggregated block.
Then we can permute 3 such blocks in 3! = 3*2*1 = 6 ways.
But we can permute the books inside each group, too, obtaining new arrangements.
It gives 4! = 24 permutations inside each group,
producing  = 6*13824 = 82944 total possible arrangements.
Thus the answer in this case is 82944 total possible arrangements.
Thus I solved the problem for two extreme cases.
Which solution to choose depends on you, my dear visitor, depending on how you understand the problem.
-----------------------
On Permutations, see introductory lessons
- Introduction to Permutations
- PROOF of the formula on the number of Permutations
- Problems on Permutations
- Arranging elements of sets containing indistinguishable elements (*)
in this site.
The most relevant is the last lesson in this list, marked by (*).
Also, you have this free of charge online textbook in ALGEBRA-II in this site
- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic "Combinatorics: Combinations and permutations".
Save the link to this textbook together with its description
Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson
into your archive and use when it is needed.
|
|
|
