Combinatorics Problem: Placing n Books with m Broken

In summary, we can place n books, of which m are broken, on a shelf so that there are at least 2 consecutive broken books. The broken books cannot be told apart, and the books in good condition are also included.
  • #1
lesdavies123
16
0
Hi, just a simple combinatorics problem I can't figure out how to do!

We want to place n books, of which m are broken, on a bookshelf so that there are at least 2 consecutive broken books. The broken books are indissociable from one another and so are the good book, how many ways can we do this

Thank you in advance!
 
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  • #2
lesdavies123 said:
Hi, just a simple combinatorics problem I can't figure out how to do!

We want to place n books, of which m are broken, on a bookshelf so that there are at least 2 consecutive broken books. The broken books are indissociable from one another and so are the good book, how many ways can we do this

Thank you in advance!
indissociable
What does this word mean? I suggest you rewrite the last sentence.
 
  • #3
mathman said:
What does this word mean? I suggest you rewrite the last sentence.

I am sorry, I am French-Canadian and didnt know how to write this word in English (thought it was the same thing :P), I meant that the broken books are all exactly the same and can't be told apart from each other and so are the books in good condition!
 
  • #4
Please clarify the original question. Do you mean any arrangement as long as at least two broken books are together? If m > (n+1)/2, all arrangements will work.
 
  • #5
Ok I will rewrite it as an exact translation of the original question: We want to place n books, of which m are broken, on a shelf so that there are at least 2 consecutive broken books. The borken books cannot be told apart one from another, and that is also the case for the books in good condition. In how many ways can we do this?

Now this is a problem given to me by my teacher's assistant who doesn't explain as well as the teacher. She gave us the answer, but I can't figure out how she got to it, this is it

n!/(m!(n-m))! - ((n-m+1) choose m) or n!/(m!(n-m))! - (n-m+1)!/(m!(n-2m+1)!)

The only thing I know is that the +1 at the end of the second term is for if we placed a broken book at the edge of the shelf because she said this in class. Otherwise, isn't n choose m (the first term) the different ways to select m books from n, but what does that signify in this case as the m books will all be on the shelf anyways? So that's that thank you in advance!
 

1. How many ways can n books be placed on a shelf with m broken spaces?

The number of ways to place n books on a shelf with m broken spaces can be calculated using the combination formula, nCr = n! / ((n-r)! * r!), where n is the total number of books and r is the number of broken spaces. This will give us the total number of combinations possible.

2. How do I know if a particular combination is valid?

In order for a combination to be valid, the number of books placed on the shelf must equal the total number of books (n) minus the number of broken spaces (m). Additionally, each book must be placed in a consecutive order without any gaps or overlaps.

3. Can I rearrange the books once they are placed on the shelf?

Yes, you can rearrange the books on the shelf as long as the combination remains valid. This means that the total number of books placed on the shelf must still equal n minus m, and the books must be placed in a consecutive order without any gaps or overlaps.

4. Is there a maximum number of broken spaces that can be accommodated?

No, there is no maximum number of broken spaces that can be accommodated. However, the more broken spaces there are, the fewer valid combinations there will be.

5. How does the number of broken spaces affect the total number of valid combinations?

The number of broken spaces affects the total number of valid combinations by reducing the total number of ways the books can be arranged. As the number of broken spaces increases, the number of valid combinations decreases.

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