Speaking of coincidences: to distract from some dentistry (hey, it worked for Pascal!) I tried to find--eyes closed--a natural way to enumerate the simple set-construction below that, surprisingly (to me), has 109 cases. How might this weird number break down? 108+1? Or 64+36+9? Or what? After deciding this puzzle might be too trivial for math-fun, I checked into a hotel and they coincidentally assigned me room 109! OK, I won't fight it. As a warm-up, for n=2 we start with a set of two elements, S2 = AB. Call the set of non-empty subsets of S2, T2 = A, B, AB. Question: how many subsets of T2 when unioned together equal S2? You should get 5: A+B, AB, AB+A, AB+B and AB+A+B Now for n=3, we have S3 = ABC, and T3 = A, B, C, AB, BC, AC, ABC. Next Question: enumerate the subsets of T3 whose union is S3. You should get 109. Can you organize these into some natural scheme? See OEIS sequence A003465 for other n.