amargius wrote:
So I'm trying to figure out the underlying logic behind a specific combinatorics problem.
"There are 10 composers and only 5 pages in the upcoming playbill. How many combinations exist if the musicians do NOT care about page order?"
Would be 10!/(5!*5!) because there are 10 pages with 5 composers in and 5 composers out (doesn't matter order so YYYYYNNNNN), so 10! over 5! times 5!.
Now what if... "There are 10 composers and only 5 pages in the upcoming playbill. How many combinations exist if the musicians DO care about page order?"
Would this be 10!/5!, because there are 10 pages with 5 the same and 5 different (12345NNNNN)?
"There are 10 composers and only 5 pages in the upcoming playbill. How many combinations exist if the musicians do NOT care about page order?"
Would be 10!/(5!*5!) because there are 10 pages with 5 composers in and 5 composers out (doesn't matter order so YYYYYNNNNN), so 10! over 5! times 5!.
Now what if... "There are 10 composers and only 5 pages in the upcoming playbill. How many combinations exist if the musicians DO care about page order?"
Would this be 10!/5!, because there are 10 pages with 5 the same and 5 different (12345NNNNN)?
From the question, this is what I gather: 10 composers - 5 pages - one composer has to compose 1 page.
When we say, 'page order does not matter', it means, we don't have page 1, page 2, page 3 etc. We just have 5 blank pages, all equal, so we just need to select 5 composers out of 10 and we will make each composer compose a page and then put them in any order.
We can do this in 10C5 ways (select 5 people out of 10) = 10!/(5!*5!) (use combinations formula - it helps you think quickly)
When the page order does matter, it means we have 5 blank pages with page numbers written on them. Now we have to select 5 composers and allot them the pages: Page 1 to composer A, page 2 to composer B etc or Page 2 to composer A, page 1 to composer B etc and so on...
Select 5 composers out of 10 as before: 10C5 = 10!/(5!*5!)
Arrange 5 pages among the 5 composers in 5! ways.
Total different cases = 10!/(5!*5!) * 5! = 10!/5!
