Counting problems: dancing couples & books and children

[member 587159]

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 :)

 

[Ray Vickson]

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?

 

[haruspex]

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?)

 

[member 587159]

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

 

[haruspex]

1) from 10 pairs one

So how many choices is that?

 

[member 587159]

So how many choices is that?

2^10

 

[haruspex]

2^10

Indeed.

 

[member 587159]

Thanks for your help :)