62. A father has five pairwise distinct oranges, which he gives to his eight sons so that each receives either one orange or nothing. In how many ways can this be done?
62. Let the number of ways to distribute 5 pairwise distinct oranges to eight sons be denoted by $A_{8}^{5}$. We will calculate $A_{8}^{5}$ using two methods.

First method. The number of ways to choose 5 out of 8 sons (those who will receive the oranges) is $C_{8}^{5}$. For any such choice, there are 5! ways to distribute the 5 pairwise distinct oranges to the 5 chosen sons. Therefore, $A_{8}^{5}=C_{8}^{5} \cdot 5!$.

Second method. The first orange can be given to any of the 8 sons. The second orange can be given to any of the remaining 7 sons, and so on. The fifth orange can be given to any of the remaining 4 sons (the other three will not receive any oranges). Thus, $A_{8}^{5}=8 \cdot 7 \cdot 6 \cdot 5 \cdot 4=\frac{8!}{3!}$.