To place 3 non-adjacent A’s in 8 positions, we use the combinatorial method of placing k non-adjacent items in n slots.

February 22, 2026

To place 3 non-adjacent A’s in 8 positions, we use the combinatorial method of placing k non-adjacent items in n slots.

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The remaining 5 segments are chosen from U, C, G (each can be used any number of times, including zero), and are distinct in labeling (but not required to be unique beyond count per type).
We proceed in two steps: first place the 3 A’s such that no two are adjacent, then assign non-A labels to the remaining positions.
Step 1: Place 3 non-adjacent A’s in 8 positions.
This is equivalent to placing 3 A’s and 5 non-A’s, with at least one non-A between any two A’s.
To count the number of ways to place 3 non-adjacent A’s in 8 positions:
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.

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