【Question 28】
Place five cards $A, K, Q, J, 10$ randomly into five envelopes labeled $A, K, Q, J, 10$, with one card in each envelope. How many ways are there to place the cards so that each card ends up in the wrong envelope?
【Analysis and Solution】
(Method One)
Consider card $A$ being placed in envelope $K$:
(1) If card $K$ is placed in envelope $A$, there are 2 scenarios:
(1) Card $Q$ is placed in envelope $J$, card $J$ is placed in envelope 10, and card 10 is placed in envelope $Q$;
(2) Card $Q$ is placed in envelope 10, card 10 is placed in envelope $J$, and card $J$ is placed in envelope $Q$;
(2) If card $K$ is placed in envelope $Q$, there are 3 scenarios:
(1) Card $Q$ is placed in envelope $A$, card $J$ is placed in envelope 10, and card 10 is placed in envelope $J$;
(2) Card $Q$ is placed in envelope $J$, card $J$ is placed in envelope 10, and card 10 is placed in envelope $A$;
(3) Card $Q$ is placed in envelope 10, card 10 is placed in envelope $J$, and card $J$ is placed in envelope $A$;

Similarly, if card $K$ is placed in envelope $J$ or 10, there are 3 scenarios each;
Therefore, if card $A$ is placed in envelope $K$, there are $2+3 \times 3=11$ scenarios;
Similarly, if card $A$ is placed in envelope $Q$, $J$, or 10, there are 11 scenarios each;
There are $11 \times 4=44$ ways to place the cards so that each card ends up in the wrong envelope.
(Method Two)
【Bernoulli - Euler Envelope Problem】
A person writes several letters and writes the corresponding addresses on several envelopes. How many ways are there to place all the letters in the wrong envelopes?
For a permutation of $n$ elements, where no element is in its correct position, there are $n!\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots(-1)^{n} \frac{1}{n!}\right)$ ways. For 5 ! $\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}\right)=44$ ways to place the cards so that each card ends up in the wrong envelope.