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)

please if u dont know question dont answer it and dont waste my time and your time!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

One way to attack this is to see how many ways 20 books can be put in a row and then see how may ways that row can be divided into 5 shelves.

For the row of 20, there are 20 choices for the first book, for each of the 20 choices of the first book there are always 19 left for the second giving 20*19 ways of picking the first two books. Extending this on there are 18 choices for the third, etc. So thee are 20*19*18*17.......*4*3*2*1or 20! ways to arrange the 20 books in order. That's 2.43 x 10^18.

Now for each ordering, there are different ways to divide the order, without shuffling, among the shelves. Since we have the books in order, we will keep them that way with the first set going in order to shelf one, next group to two, etc. That way, we only need to figure how many ways the 20 books can be divided without regard to which book is which. Each ordering above can go into any of the schemes to divide the books among the shelves so you just multiply the two numbers together to get the total number of possibilities.

Since each shelf must have two books minimum, start by putting two books on each shelf and leaving them there. The problem then comes down to how many ways can the remaining 10 books be divided among 5 shelves. We really want to put 4 divisions along the row of 10 books to tell where one shelf ends and the other begins. This gives 14 locations to be occupied by either a book or divider. The number of possible arrangements of the 14 locations is 14!. But since all dividers look the same and all books (for now) look the same, there are only 14!/(10! * 4!) = 1001 possible ways to divide it onto the shelves.

So there are 1001 ways to divide the 20 books among the shelves for each of the 20! Arrangements of books. Multiplying these together gives 1001*20! = 2.43 x 10^21 ways to arrange the books. If you want to see all the arrangements, I suggest you start arranging books right away. If you put together one arrangement of books per second it would take about 5000 times the age of the universe to go through all of them.

Answer 2

Are you in a discrete math class? I hate questions like this...but I'm going to have to answer them on a final next Tuesday, so this is good for me too.

Here's what I think. See if you agree.

First, lets look at the problem assuming that the books were all the same.

Then you'd be forced to use 10 of the books first, to guarantee that there were 2 on each shelf. You'd then have 10 books left to place on the five shelves.

This is where it gets a little tricky.

You could list out every possibility, but there are a lot of them. For instance, you could put all 10 on one shelf 4 different ways. You could put 9 on one shelf 4 different ways, and 1 on one of the remaining 3 shelves, so that's 12 more possibilities. Etc etc.

But (borrowing from my discrete math book's technique), you can also think about drawing a diagram of the possibilities, using "stars" and "bars". The stars represent the books, and the bars indicate where one shelf ends and the next begins.

So, for instance, the arrangement "3 books on the top 2 shelves, and 2 on the bottom 2" would look like:

* * * | * * * | * * | * *

where the first three *s represent the 3 books on the top shelf, the | means go to the next shelf, etc. Hope that makes sense.

Now, realize that every possible diagram will consist of 10 stars and 3 bars, for a total of 13 possible "locations":

_ _ _ _ _ _ _ _ _ _ _ _ _

10 of the positions have to be filled with stars, and the other 3 will be filled with bars. So (here I'm assuming you're familiar with "combinations"), there are C(13, 10) ways to do that, since order does not make a difference. C(13, 10) = C(13, 3) = 13 x 12 x 11 / (3 x 2 x 1) = 858 ways to arrange the remaining 10 books over the other 4 shelves.

That would be the final answer if all the books were the same. But the problem says that all the books are different. So, for each of the 858 possible arrangements of books, you would have 20 choices for the location of the first book, 19 for the second, 18 for the third, etc etc. So the final answer is actually 858 x 20! = 858 x 2,432,902,008,176,640,000 = 2,087,429,923,015,557,120,000.

That's a huge number, but I'm pretty sure it's correct.

Answer 3

The answer is 20 factorial ways because this is the combination of 20 things taking 20 of them at a time. Arrange them atleast 2 in one shelf or more which ever way you please it won't make any difference to the number of combinations. Once you get the series from different combinations it is up to you to decide how to arrange them starting from the first shelf to the last in which ever number you decide to place in each shelf. There is no complication.

Answer 4

go back to the last time you asked the question .... a guy called Hy and I worked it out and showed the variasions (just add the basic 2 books on every shelf)
the answer is 971 ways