We use the stars and bars transformation: if we place 3 A’s with at least one non-A between them, we first reserve 2 non-A’s to enforce spacing. So, we place 3 A’s and use 2 of the 5 non-A’s as "separators" between them. That leaves $5 - 2 = 3$ non-A’s to distribute freely in the 4 possible gaps: before the first A, between A1 and A2, between A2 and A3, and after the third A.

Sun, 10 May 2026 09:06:36 +0000

Content under review