1) We choose randomly 10 persons from 10 dance couples, but we need at least one dance couple. In how many ways is this possible?
2) If we divide n+1 books among n children and every child has to get at least one, in how many ways would this be possible?
Combinations, permutations, variations
The Attempt at a Solution
1) We want to choose 10 people from 20 people, this is possible in 20 nCr 10 ways.
We need a couple so we need less than 20 nCr 10 ways. I want to know then:
20 nCr 10 - the amount of possibilities to have no partner. This is where I got stuck.
2) We can give n+1 books to n students in (n+1) nPr n ways, but once every pupil has got a book, there is one book remaining. We can give this to any of the students in n ways, but since it doesn't matter in what order this student received the book, we should multiply with 1/2.
So,[(n+1) nPr n]*n/2 ways
Can someone verify this?
Thanks in advance for your quick answers :)