Each of the $5$ people leaves a coat and a hat in the cloakroom. The absent-minded cloakroom attendant gives each person a random coat and a random hat. Calculate the probability of the event that each of the $5$ people gets someone else's coat or someone else's hat.
To count the good cases, I thought to subtract from all possibilities those where at least one person gets both their own coat and their own hat. I used the principle of inclusion and exclusion.
Initially, we assume that one person has their things, so we have $|A_i| = \binom{5}{1} \cdot 4! \cdot 4!$, where we choose the person with $\binom{5}{1}$, and we assign the remaining hats and coats with $4! \cdot 4!$. For two people it is $\binom{5}{2} \cdot 3! \cdot 3!$, etc.
So my solution is
$$\frac{1}{n! \cdot n!}\sum_{i=0}^n (-1)^i \color{blue}{\binom{n}{i}} ((n-i)!)^2 = \frac{11844}{14400} \approx 0.82$$
And in the answers, there is
$$\frac{1}{n! \cdot n!}\sum_{i=0}^n (-1)^i ((n-i)!)^2 = \frac{13856}{14400} \approx 0.96$$
Why don't we want to choose the person who will get their things?
Best Answer
The probability $P$ that each person gets their own coat & hat is equal to $c \cdot h$ where c is the probability of each person getting their own coat and h is the probability of each person getting their own hat. Since these are independent variables, but are evaluated the same way, we can find that $P=q^2$. To find $q$, we can observe that the first person has a $1/5$ probability of getting the correct article, the second $1/4$, and so on. Ultimately, we find that for $n$ people, $q=1/n!$. Since $q^2=P$, $P(n)=\frac{1}{(n!)^2}$. Evaluating at $5$, we get $\frac{1}{(1 \cdot 2 \cdot 3 \cdot 4 \cdot 5)^2}=\frac{1}{(120)^2}=\frac{1}{1400}$.