*From Gardner's Colossal book of short puzzles and problems* >A secretary types four letters to four people and addresses the four envelopes. If she inserts the letters at random, what is the probability that exactly three letters will go into the right envelopes? It's 0. If three letters go into the right envelope, then the remaining letter can't be anywhere but in the right envelope as well. Picking the first letter at random, there's a $\frac{1}{4}$ probability for it to end up in the right envelope. For the second, it's $\frac{1}{3}$, because only 3 letters remain. For the third, it's $\frac{1}{2}$. But the question asks about the probability of exactly three letters going into the right envelope. This means the last probability by which we must multiply is $\frac{0}{1}$, because out of the 1 letters remaining, there's 0 chance for it to end up in the wrong envelop if all other letters are in the right one. $ \begin{align} P(\text{Three letters are in the right envelope}) \\[2mm] =\frac{1}{4} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{0}{1} \\[2mm] =0 \end{align} $ Using the combinations method, $ \begin{align} P(A)&=\frac{\text{favorable outcomes}}{\text{total outcomes}} \\[2mm] &=\frac{0}{4!} \\[2mm] &= 0 \end{align} $