HMMT 十一月 2017 · 冲刺赛 · 第 27 题
HMMT November 2017 — Guts Round — Problem 27
题目详情
- [ 15 ] On a 3 × 3 chessboard, each square contains a Chinese knight with probability. What is the 2 probability that there are two Chinese knights that can attack each other? (In Chinese chess, a Chinese knight can attack any piece which is two squares away from it in a particular direction and one square away in a perpendicular direction, under the condition that there is no other piece immediately adjacent to it in the first direction.) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2017, November 11, 2017 — GUTS ROUND Organization Team Team ID#
解析
- [ 15 ] On a 3 × 3 chessboard, each square contains a Chinese knight with probability. What is the 2 probability that there are two Chinese knights that can attack each other? (In Chinese chess, a Chinese knight can attack any piece which is two squares away from it in a particular direction and one square away in a perpendicular direction, under the condition that there is no other piece immediately adjacent to it in the first direction.) Proposed by: Yuan Yao 79 Answer: 256 A B C D E F Suppose the 3 × 3 square is We count the number of ways a board could have two knights G H I attack each other using PIE. First notice that in any setup with two knights attack each other, the center square must be empty. Also, for any pair of knights that attack each other, one must be in a 5 corner, and the other at the center of a nonadjacent side. There are 8 · 2 ways for one pair of knights to attack each other. Next, we count the number of ways two pairs of knights attack each other: up to symmetry, there are four cases: knights at A, B, G, H, and D and E empty; knights at A, H, F, and B, D, E empty; knights at A, B, H, I, and D, E, F empty; and knights at A, C, H, and D, E, F empty. 3 3 2 3 Four each of these cases, there are four symmetries, so there are a total of 4 · (2 + 2 + 2 + 2 ) ways to have two pairs of knights attack each other. Next, there’s only one way for three pairs of knights to attack each other, discounting symmetry: A, B, G, H, I have knights, and D, E, F empty. Then there are 4 · 2 · 2 ways for three knights to attack. Finally, there is only one way for four knights to attack: knights at A, B, C, G, H, I and empty squares at D, E, F, for a total of 2 ways after counting symmetries. Applying PIE, we get that the total number of boards with at least one pair of knights attacking each other is 5 3 3 2 3 8 · 2 − 4 · (2 + 2 + 2 + 2 ) + 4 · 2 · 2 − 2 = 158 . 79 158 Then the probability the 3 × 3 board has a pair of knights attacking each other is = . 9 2 256