Counting problems: dancing couples & books and children

  • Thread starter member 587159
  • Start date
  • #1

Homework Statement



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?

Homework Equations



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 :)
 

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

Homework Statement



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?

Homework Equations



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 :)

Is a dancing couple any man paired with any woman, or is it a particular pairing such as Fred and Ethel?
 
  • #3
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
38,831
8,219
the amount of possibilities to have no partner
If you do not choose both of any pair, how many pairs will you choose one from?
[(n+1) nPr n]*n/2 ways
Yes, but can you simplify that, avoiding the use of nPr. (Also, I don't think this is standard notation. Don't you just mean [(n+1) P n]*n/2?)
 
  • #4
If you do not choose both of any pair, how many pairs will you choose one from?

Yes, but can you simplify that, avoiding the use of nPr. (Also, I don't think this is standard notation. Don't you just mean [(n+1) P n]*n/2?)

1) from 10 pairs one
2) this is the notation from my graphing calculator, but it's not the standard notation

Thanks for your reply
 
  • #8
Thanks for your help :)
 

Suggested for: Counting problems: dancing couples & books and children

Replies
7
Views
543
Replies
14
Views
601
Replies
1
Views
478
Replies
2
Views
447
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
4
Views
760
Replies
1
Views
273
Replies
5
Views
296
  • Last Post
Replies
2
Views
418
Top