Four letters have been written to four different people and the four addressed envelopes are lying ready. A clerk in a hurry pushes one letter into each envelope at random. What is the probability that exactly two of the letters end up in their correct envelopes?

The letters can be placed in 4! = 24 ways, all equally likely. For exactly two to be right, first choose WHICH two are correct: $C(4,2)=6$ ways. The remaining two letters must then BOTH be wrong, and two objects can be misplaced in only one way — each goes into the other one’s envelope. So the favourable count is $6\times 1=6$, and the probability is 6/24 = 1/4. Trap 1/2 = 12/24 comes from multiplying by 2! for the last two letters; that second arrangement puts them both right, which would make all four correct, not two. Trap 1/6 = 4/24 uses 4 rather than $C(4,2)=6$ for choosing the correct pair. Trap 1/12 = 2/24 counts only the two adjacent pairs. As a check, the four cases are: none right 9, exactly one 8, exactly two 6, all four 1 — and 9 + 8 + 6 + 1 = 24, with exactly three right being impossible.