Back to HW 4: 5.5
Could someone help me out with this one?
Examine the situation as follows:
We must place the books we will choose
| | | | |
Now we must place at least one book between each of these to separate them
|*|*|*|*|
We have so far placed the five books we will choose, and four books to separate them. The remaining three books can either be placed next to the separating books, or on the outsides of the current model we have constructed. We now have six places to place three books. --User:mturczi
Would there not be more choices than that, because you can place the 3 remaining books in between any of the 9 books on the shelf? --User:jdrummon
I'm not sure what I am to do with this problem.. Is it something similar to the arranging women and men, and two men cannot be arranged in a row...? --- To add to what the original poster mentioned:
We know that we have 5 "chosen books (notated by | )" and at least 4 books (notated by *)used to separate the chosen ones: |*|*|*|*| We then can see the total amount of places that the remaining 3 books can be placed, (notated by _ ) _|_*_|_*_|_*_|_*_|_ As you can see, putting the remaining 3 books into any of these "spaces" will still retain the integrity of the problem. --Therefore, we need to find the number of ways to place 3 books into 10 spaces.
We do this like we have done several problems before:
10 categories, (9 bars, each separating the categories):
| | | | | | | | |
We need to place 3 books, so using what we've learned we take
("number of separators" + "number of items to be placed") choose ("number of items to be placed")
So,
9+3 choose 3 12 choose 3
= 12!/(3!)(9!) =220
hope this helps
--Msstaffo 12:12, 12 February 2009 (UTC)
