number of ways to partition $S=\{1,2,\dots,2020\}$ so that given $a\in A, b\in B: \ a + b $ is never a multiple of 20 and 1024 Find the number of ways to partition $S=\{1,2,\dots,2020\}$ into two disjoint sets A and B with $A \cup B = S$ so that if you choose an element $a \in A$ and $b \in B:\  a + b$ is never a multiple of 20. A or B can be the empty set, and the order of A and B doesn’t matter. In other words, the pair of sets (A, B) is indistinguishable from the pair of sets (B, A).
my first approach was to try to work out for a simpler case like $S=\{1,2,\dots,20\}$. I noted $20=(19+1)=(18+2)=(17+3)= \dots =(9+11)$, the interesting thing about these pairs is that notice that the elements of each pair should strictly be in the same set otherwise would add up to yield a multiple of 20, the remaining were 20 and 10 out of which 10 could be placed in either group and the element 20 could also go in the either set except in the case where we consider null sets.
according to me in the simpler case discussed above, if i left out the null sets and the hassle with 10 and 20, the total number of such required sets i was getting as 9C1+9C2+9C3+ . . . + 9C8
 A: First, assume we have an acceptable partition and let $x \equiv y \pmod{20}$.  If $x$ and $y$ are in different sides of the partition, then at least one of them must be in the same side as $20-x$.  That contradicts our assumption that the partition is acceptable.  Thus, all elements of the same equivalence class $\pmod {20}$ must go within the same side of the partition.
That means that the simpler case you have already analyzed gives us our solution.  There are nine pairs of equivalence classes that add to $20$.  Each must go into the same side of the partition.  In addition, the equivalence class of $10$ and the equivalence class of $20$ can go into either side of the partition.
Let's arbitrarily designate the side of the partition containing the equivalence class of $20$ as $A$.  Then you have $2$ choices for each of the remaining $10$ "pairs" of equivalence classes (where we treat the equivalence class of $10$ as a pair).  Each "pair" can either go into $A$ or into $B$.  Thus, the total number of acceptable partitions is $2^{10}=1024$.
