PUMaC 2020 · 组合（B 组） · 第 2 题

PUMaC 2020 — Combinatorics (Division B) — Problem 2

Joey is playing with a 2-by-2 Rubik’s cube made up of 8 1-by-1 cubes (with 2 of these smaller cubes along each of the sides of the bigger cubes). Each face of the Rubik’s cube is distinct color. However, one day, Joey accidentally breaks the cube! He decides to put the cube back together into its solved state, placing each of the pieces one by one. However, due to the nature of the cube, he is only able to put in a cube if it is adjacent to a cube he already placed. If different orderings of the ways he chooses the cubes are considered distinct, determine the number of ways he can reassemble the cube.

解析

Joey is playing with a 2-by-2-by-2 Rubik’s cube made up of 8 1-by-1-by-1 cubes (with two of these smaller cubes along each of the sides of the bigger cubes). Each face of the Rubik’s cube is distinct color. However, one day, Joey accidentally breaks the cube! He decides to put the cube back together into its solved state, placing each of the unit cube pieces back one by one. However, due to the nature of the cube, he is only able to put in a unit cube if it is adjacent to a unit cube he already placed. If different orderings of the ways he chooses the cubes are considered distinct, determine the number of ways he can reassemble the cube. Proposed by Frank Lu Answer: 8640 . Solution: We have 8 choices for the first cube that Joey picks up. Then, he has 3 choices for the second cube and 4 for the third cube, yielding us 96 ways to first construct an L made up of three cubes. Now, note that there are 4 places to put the fourth cube. If Joey decides to not place the cube on top of the center of the L, then we can observe that each of the 4 remaining spots can be filled in any order. Otherwise, if Joey places the cube on the top of the center of the L, he has 3 places for the next cube, and then he can fix the cube with the last 3 pieces in any order. This yields 96 · (3 · 24 + 18) = 96 · 90 = 8640 ways that Joey can fix the cube.