# Counting...

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• May 4th 2013, 12:03 PM
Leviathantheesper
Counting...
Well, hi.

I've got a problem, I've been trying to solve it but every formula I try to make fails...

Perhaps this is foolish and I'm not applying correctly the Permutations, Combinations and Factorials...but I don't get to the answer.

The problem is:

Let's Suppose that I'm a teacher, I take some notebooks from nine students, and I give them back. ¿How many ways are there to give back the notebooks in a way that no one receives his own notebook?

What I do, naturally, is to reduce the problem, and try making a formula.

I look what happens with four notebooks...I started looking how many ways there are...and I find there are 9 ways. With 3 notebooks there are only 2 ways. And with 2 there is only 1. No easy-looking pattern, so I can't make a formula that easy...

Then I tried using a formula that I already know...the total quantity of ways to give back the notebooks is n!. Then the formula must be
Attachment 28244.

When I try to use all of that I get a weird +...-... thing like n!-n!+(n-1)!-...

But it's not exactly the formula...and I'm not sure if it really can be done like that...
• May 4th 2013, 12:24 PM
Plato
Re: Counting...
Quote:

Originally Posted by Leviathantheesper
The problem is:
[FONT=arial black]
Let's Suppose that [FONT=arial]I'm a teacher, I take some notebooks from nine students, and I give them back. ¿How many ways are there to give back the notebooks in a way that no one receives his own notebook?
What I do, naturally, is to reduce the problem, and try making a formula.
I look what happens with four notebooks...I started looking how many ways there are...and I find there are 9 ways. With 3 notebooks there are only 2 ways. And with 2 there is only 1. No easy-looking pattern, so I can't make a formula that easy...

This is known as the problem of derangement.
I don't like the notation there, but it gives you the ideas.

I like the formula: $D(n) = (n!)\sum\limits_{k = 0}^n {\frac{{{{( - 1)}^{k }}}}{{k!}}} \approx \frac{{n!}}{e}$.