# Counting problems: dancing couples & books and children

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2019 Award

## 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?

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Ray Vickson
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Dearly Missed

## 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?

Is a dancing couple any man paired with any woman, or is it a particular pairing such as Fred and Ethel?

haruspex
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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?)

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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

haruspex
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1) from 10 pairs one
So how many choices is that?

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So how many choices is that?
2^10

haruspex
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2^10
Indeed.

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