Santa Claus arrived at Arnaldo and Bernaldo's house carrying ten distinct toys numbered from 1 to 10 and said to them: "the toy number 1 is for you, Arnaldo and the toy number 2 is for you, Bernaldo. But this year, you can choose to keep more toys as long as you leave at least one for me". In how many ways can Arnaldo and Bernaldo divide the remaining toys between them?
Solution

For each of the 8 toys, from number 3 to number 10, we must decide whether it belongs to Arnaldo, Bernaldo, or should be left for Santa Claus. If we multiply then

$$
\underbrace{3 \times 3 \times \cdots \times 3}_{8 \text { times }}
$$

we will count the ways to distribute the toys among Arnaldo, Bernaldo, and Santa Claus, including the cases where Santa Claus gets no toys. It remains to count the number of ways to distribute all the toys between Arnaldo and Bernaldo (without leaving anything for Santa Claus), and subtract this number from $3^{8}$.

To distribute the toys between Arnaldo and Bernaldo, we must decide, for each of the 8 toys, which of the two will get the toy. Thus, we have

$$
\underbrace{2 \times 2 \times \cdots \times 2}_{8 \text { times }}
$$

ways to distribute the toys between Arnaldo and Bernaldo. Finally, the answer is

$$
3^{8}-2^{8}=6305
$$

ways to distribute the toys.