Problem source

Starting with the result $\P(A\cup B) = \P(A)+P(B) - \P(A\cap B)$, prove that 
\[
\P(A\cup B\cup C) = \P(A)+\P(B)+\P(C)  
- \P(A\cap B) - \P(B\cap C) - \P(C \cap A)
+ \P(A\cap B\cap C)
\,.
\] 
Write down, without proof, the  corresponding result for four events $A$, $B$, $C$ and $D$.
A pack of $n$ cards, numbered $1, 2, \ldots, n$, is shuffled and laid out in a row. The result of the shuffle is that each card is equally likely to be in any position in the row. Let $E_i$ be the event that the card bearing the number $i$  is in the $i$th position in the row. Write down the following probabilities:
\begin{questionparts}
\item $\P(E_i)$;
\item $\P(E_i\cap E_j)$, where $i\ne j$;
\item  $\P(E_i\cap E_j\cap E_k)$, where $i\ne j$, $j\ne k$ and $k\ne i$.
\end{questionparts} 
Hence show that the probability that at least one card is in the same position as the number it bears is
\[
1 - \frac 1 {2!} + \frac 1{3!} - \cdots + (-1)^{n+1} \frac 1 {n!}\,.
\]
Find the probability that exactly one card is in the same position as the number it bears 

Solution source

\begin{align*}
&& \mathbb{P}(A \cup B \cup C) &= \mathbb{P}(A \cup B) + \mathbb{P}(C) - \mathbb{P}((A \cup B) \cap C) \tag{applying with $A\cup B$ and $C$} \\
&&&=  \mathbb{P}(A \cup B) + \mathbb{P}(C) - \mathbb{P}((A \cap C) \cup (B \cap C)) \\
&&&=  \mathbb{P}(A)+\mathbb{P}(B) - \mathbb{P}(A\cap B) + \mathbb{P}(C) - \mathbb{P}((A \cap C) \cup (B \cap C))  \tag{applying with $A$ and $B$}\\
&&&=  \mathbb{P}(A)+\mathbb{P}(B) - \mathbb{P}(A\cap B) + \mathbb{P}(C) - \left ( \mathbb{P}(A \cap C) +\mathbb{P}(B \cap C) - \mathbb{P}( (A \cap C) \cap (B \cap C) )\right) \\
&&&= \mathbb{P}(A)+\mathbb{P}(B) +\mathbb{P}(C)- \mathbb{P}(A\cap B)- \mathbb{P}(A \cap C) -\mathbb{P}(B \cap C)+\mathbb{P}( A \cap B \cap C)
\end{align*}

\[ \mathbb{P}(A_1 \cup A_2 \cup A_3 \cup A_4) = \sum_i \mathbb{P}(A_i) -  \sum_{i \neq j} \mathbb{P}(A_i \cap A_j) + \sum_{i \neq j \neq j} \mathbb{P}(A_i \cap A_j \cap A_k) - \mathbb{P}(A_1 \cap A_2 \cap A_3 \cap A_4) \]

\begin{questionparts}
\item $\mathbb{P}(E_i) = \frac{1}{n}$
\item $\mathbb{P}(E_i \cap E_j) = \frac{1}{n} \cdot \frac{1}{n-1} = \frac{1}{n(n-1)}$ 
\item $\mathbb{P})(E_i \cap E_j \cap E_k) = \frac{1}{n(n-1)(n-2)}$
\end{questionparts}

First notice that the probability that $k$ (or more) cards are in the correct place is $\frac{(n-k)!}{n!}$ (place the other $n-k$ cards in any order. We are interested in:

\begin{align*}
\mathbb{P} \left ( \bigcup_{i=1}^n E_i \right) &= \sum_{i} \mathbb{P}(E_i) - \sum_{i \neq j} \mathbb{P}(E_i \cap E_j) +  \sum_{i \neq j \neq k} \mathbb{P}(E_i \cap E_j \cap E_k) - \cdots  \\
&= \sum_i \frac1n - \sum_{i\neq j} \frac{1}{n(n-1)} + \sum_{i \neq j \neq k} \frac{1}{n(n-1)(n-2)} -  \cdots  + (-1)^{k+1} \sum_{i_1 \neq i_2 \neq \cdots \neq i_k} \frac{(n-k)!}{n!} + \cdots\\
&= 1 - \binom{n}{2} \frac{1}{n(n-1)} + \binom{n}{3} \frac{1}{n(n-1)(n-2)} - \cdots + (-1)^{k+1} \binom{n}{k} \frac{(n-k)}{n!} + \cdots \\
&= 1 - \frac12 + \frac1{3!} - \cdots + (-1)^{k+1} \frac{n!}{k!(n-k)!} \frac{(n-k)!}{n!} + \cdots \\
&= 1 - \frac1{2!} + \frac1{3!} - \cdots + (-1)^{k+1} \frac{1}{k!} + \cdots + (-1)^{n+1} \frac{1}{n!} 
\end{align*}

The probability exactly one card is in the right place is the probability none of the other $n-1$ are in the right place, which is:

$\frac1n \left (1 - \left (1 - \frac1{2!} + \frac1{3!} - \cdots + (-1)^{k+1} \frac{1}{k!} + \cdots + (-1)^{n} \frac{1}{(n-1)!}  \right) \right)$

but there are also $n$ cards we can choose to be the card in the right place, hence

$\frac{1}{2!} - \frac{1}{3!} + \cdots +(-1)^n \frac{1}{(n-1)!}$