6.1. On 19 cards, the numbers $15,16,17, \ldots, 33$ are written respectively (one number per card). Members of the math club Vasya, Petya, and Misha decided to divide all these cards among themselves so that each of them gets at least one card and no one ends up with a pair of cards where the difference between the numbers is odd. How many ways are there to distribute the cards?
Answer: 4596.

Solution. According to the condition, each participant will have either only even-numbered cards or only odd-numbered cards. We choose a participant (3 ways), and give them all 9 cards with even numbers. The remaining 10 cards with odd numbers are distributed between the two others, which can be done in $2^{10}$ ways. However, there will be 2 ways in which one of these two participants ends up without any cards. Therefore, we have $3 \cdot\left(2^{10}-2\right)$ ways.

Similarly, we give all the cards with odd numbers to one participant, and distribute the remaining 9 cards between the other two. Here, we have $3 \cdot\left(2^{9}-2\right)$ ways.

In total, the number of ways is: $3 \cdot\left(2^{10}+2^{9}-4\right)=3 \cdot 4 \cdot\left(2^{8}+2^{7}-1\right)=4596$.