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Home » Thus, the final answer is $oxed3$. - Nelissen Grade advocaten

Thus, the final answer is $oxed3$. - Nelissen Grade advocaten

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Alternative ( clearer) approach:

📌 The number of ways to choose 3 non-adjacent positions out of 8 is a known result. It is equivalent to choosing 3 positions such that no two are consecutive.

📌 This is given by:

📌 \binom{8 - 3 + 1}{3} = \binom{6}{3} = 20

📌 This formula counts the number of ways to place $k$ non-adjacent indistinct items in $n$ positions.

📌 So, there are 20 ways to choose positions for the 3 A’s such that no two are adjacent.

📌 Step 2: Assign labels to the remaining 5 positions.

📌 Each of the 5 non-A positions can be labeled U, C, or G (3 choices each). So, number of labelings is $3^5 = 243$.

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