2016 AIME II Problem 12

Below is the professionally curated solution for Problem 12 of the 2016 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AIME II solutions, or check the answer key.

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Concepts:graph theoryrecursive counting

Difficulty rating: 2400

12.

The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.

Solution:

Let PnP_n be the number of valid paintings of a ring of nn sections. Cutting a ring open between two adjacent sections shows that ring paintings correspond exactly to rows of nn sections with adjacent colors different and the two end colors different. A row of nn sections with adjacent colors different can be painted in 43n14 \cdot 3^{n-1} ways, and the rows whose end colors match correspond, by merging the two end sections into one, to ring paintings of n1n - 1 sections. Hence Pn+Pn1=43n1.P_n + P_{n-1} = 4 \cdot 3^{n-1}.

Three mutually adjacent sections give P3=432=24,P_3 = 4 \cdot 3 \cdot 2 = 24, so P4=10824=84,P_4 = 108 - 24 = 84, then P5=32484=240,P_5 = 324 - 84 = 240, and finally P6=972240=732.P_6 = 972 - 240 = 732.

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