Probability that at least 1 person receives its letter for a distribution of n letter to n people (paired 1 to 1) I'm trying to solve an elementary probability problem, but I don't get the right answer and can't find the flaw in my reasoning. The problem and my (wrong) solution goes as follows.
Somebody distributes randomly $n$ letters to $n$ people, where each person has one letter to which it is addressed. What is the probability that at least one person gets its letter?
Let $E$ be the event such that at least one person gets the right letter. The probability space is $\Omega$. I'll denote cardinality as $\#$. The probability of $E$ is
$$\mathbb{P}(E) = \frac{\#E}{\#\Omega} $$
Obviously, $\#\Omega = n!$ .
To find $E$, I divide it into smaller independent events $A_i$, such that $A_i$ is "exactly $i$ people get the right letter". I thus have 
$$\mathbb{P}(E) = \mathbb{P}(\bigcup_{i}A_i) = \sum_{i=1}^{n} \mathbb{P}(A_i) = \sum_{i=1}^{n} \frac{\# A_i}{n!}$$
I find that $\#A_i$ is the number of unordered combinations of $i$ from $n$, that is $\frac{n!}{i!(n-i)!}$ . Replacing that in the preceding equation gives me
$$\mathbb{P}(E) = \sum_{i=1}^{n} \frac{1}{i!(n-i)!}$$
The answer is supposed to be $\mathbb{P}(E) = \sum_{i=1}^{n} \frac{(-1)^{i+1}}{i!}$ . What is the flaw in my reasoning? 
Thank you for your help!
 A: The probability that at least one person gets his/her letter is $1$ minus the probability that we have a derangement. 
There are $n!$ possible distributions of the $n$ letters to the $n$ people. 
If $D(n)$ is the number of derangements of a set of $n$ elements, then the probability $P(n)$ that at least one of the $n$ people does not get his/her letter is $$P(n) = 1 - \frac{D(n)}{n!}$$
According to Wikipedia, $$D(n) = n! \displaystyle\sum\limits_{i=0}^{n} \frac{\left(-1 \right)^{i}}{i!}$$
$$D(n) = \left\lfloor \frac{n!}{e} + \frac{1}{2} \right\rfloor $$
The second formula for $D(n)$ is far cooler, but the first one solves your problem. 
$$P(n) = 1 - \frac{n! \displaystyle\sum\limits_{i=0}^{n} \frac{\left(-1 \right)^{i}}{i!}}{n!}$$
$$P(n) = 1 - \displaystyle\sum\limits_{i=0}^{n} \frac{\left(-1 \right)^{i}}{i!}$$
We can cancel the $1$s and reindex to get the form you want:
$$P(n) = \displaystyle\sum\limits_{i=1}^{n} \frac{\left(-1 \right)^{i+1}}{i!}$$
A: For future reference, here is what I did wrong.
The events $A_{i}$ are not independent. If (n-1) people get the right letter, then it is sure that the last person also gets his letter.
