7. In "Dragon Poker," the deck has four suits. An Ace brings 1 point, a Jack -2 points, a Two $-2^{2}$, a Three $-2^{3}$, ..., a Ten $-2^{10}=1024$ points. Kings and Queens are absent. You can choose any number of cards from the deck. In how many ways can you score 2018 points?
Answer: $C_{2021}^{3}=1373734330$.

Solution. In any suit, you can score any number of points from 0 to 2047, and this can be done in a unique way. This can be done as follows: write down this number in binary and select the cards corresponding to the binary positions containing 1 (i.e., if there is a 1 in the $k$-th position, then you should take the card that brings $2^{k}$ points).

Thus, the problem reduces to partitioning the number 2018 into 4 non-negative integer addends. Reasoning similarly to the previous problem, we get the desired number of ways: $C_{2021}^{3}=1373734330$.