To place 3 non-consecutive positions: let the positions be $ x_1 < x_2 < x_3 $, with $ x_{i+1} \ge x_i + 2 $. Let $ y_i = x_i - (i-1) $, then $ y_1 < y_2 < y_3 $ in $ \{1,\dots,6\} $. So number is $ \binom{6}{3} = 20 $.

February 22, 2026

$To place 3 non-consecutive positions: let the positions be $ x_1 < x_2 < x_3 $, with $ x_{i+1} \ge x_i + 2 $. Let $ y_i = x_i - (i-1) $, then $ y_1 < y_2 < y_3 $ in $ \{1,\dots,6\} $. So number is $ \binom{6}{3} = 20 $.$

Let’s suppose we place the **C’s** with no two adjacent.
To place 3 C’s with no two adjacent in 8 positions: we first place 3 C’s such that there is at least one space between them.
The number of ways to choose 3 non-adjacent positions in 8 is a classic stars and bars adjustment.
For each such choice of 3 non-adjacent positions, we place 3 non-adjacent C’s.
Now, from the remaining 5 positions, we must place 2 A’s and 3 G’s (since 3 C’s are placed), **with no two A’s adjacent and no two G’s adjacent** — but wait: the condition is **no two identical letters adjacent**, so A’s cannot be adjacent to A, G’s to G.
But A and G are different, so A and G can be adjacent.

To place 3 non-consecutive positions: let the positions be $ x_1 < x_2 < x_3 $, with $ x_{i+1} \ge x_i + 2 $. Let $ y_i = x_i - (i-1) $, then $ y_1 < y_2 < y_3 $ in $ \{1,\dots,6\} $. So number is $ \binom{6}{3} = 20 $.

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