How many ways can you place 20 different books on5 shelve sothere is at least two bookds on each shelve?

Question

for any of these arrangemenets,consider the books on each shelf to be place one next to th other,with first book ar the left of the shelf)

Answer 1

First, find out how many ways you can arrange 20 different books with exactly two books on each shelf. Then, find how many ways you can arrange the remaining 10 books on the 5 shelves. Multiply these two answers together to find the total number of arrangements.

For the first, consider each spot on each shelf as a different position. How many choices are there for the left-most spot on the top shelf? How many choices for the second spot on the top shelf? Continue for each shelf and multiply them together - you should get 20!/10!

For the second, you have 10 books and 5 shelves where each book can go. A shelf can have no books at all or it might have all 10 books. If you start at the top shelf, they can either place a book or move to the next shelf. You have 10 books and will move to the next shelf a total of 4 times. Pretend you are putting the books in a row and using a marker to indicate when you move to the next shelf - you have 14 spots to place either a book or a marker and the markers all look alike. (This is similar to arranging the letters AAAABCDEFGHIJK) The number of ways you can arrange them is 14! / 4!

Thus, the total number of ways to arrange the books is 20!14! / 10!4!

Answer 2

To know the answer we have to use multinomial theorem:

Let (x)^r denote r books
then each shelve can have more than (x)^2
but we have a limit of 20 books on total books

so the no of ways in which we can place them are

coeff of x^20 in the eqn (x^2 + x^3 + ................+x^(inf))^5

ie the coeff of x^10 in (1 + x^1 +x^2 +x^3 ...............x^(inf))^5

ie 14c10 = 1001
if u have problem understanding it then u first have to know multinomial theory

Answer 3

What's the max number of books each shelf can hold? or would that make a difference?

Answer 4

I still think my answer is correct on the other version of this question:

http://answers.yahoo.com/question/index;_ylt=AnW5NaL7.gI40My5gA.1W3nsy6IX?qid=20061122204847AAnIoAS