HMMT 二月 2010 · 冲刺赛 · 第 28 题
HMMT February 2010 — Guts Round — Problem 28
题目详情
- [ 18 ] Danielle Bellatrix Robinson is organizing a poker tournament with 9 people. The tournament will have 4 rounds, and in each round the 9 players are split into 3 groups of 3. During the tournament, each player plays every other player exactly once. How many different ways can Danielle divide the 9 people into three groups in each round to satisfy these requirements?
解析
- [ 18 ] Danielle Bellatrix Robinson is organizing a poker tournament with 9 people. The tournament will have 4 rounds, and in each round the 9 players are split into 3 groups of 3. During the tournament, each player plays every other player exactly once. How many different ways can Danielle divide the 9 people into three groups in each round to satisfy these requirements? 9 6 3 ( )( )( ) 3 3 3 Answer: 20160 We first split the 9 people up arbitrarily into groups of 3. There are = 280 3! ways of doing this. Without loss of generality, label the people 1 through 9 so that the first round groups are { 1 , 2 , 3 } , { 4 , 5 , 6 } , and { 7 , 8 , 9 } . We will use this numbering to count the number of ways DBR can divide the 9 players in rounds 2, 3, and 4. Guts Round In round 2, because players 1, 2, and 3 are together in the first round, they must be in separate groups, and likewise for { 4 , 5 , 6 } and { 7 , 8 , 9 } . Disregarding ordering of the three groups in a single round, we will first place 1, 2, and 3 into their groups, then count the number of ways to place { 4 , 5 , 6 } and { 7 , 8 , 9 } in the groups with them. We do this by placing one member from each of { 4 , 5 , 6 } and { 7 , 8 , 9 } 2 into each group. There are (3!) ways to do this. Now, because of symmetry, we can use the round 2 grouping { 1 , 4 , 7 } , { 2 , 5 , 8 } , { 3 , 6 , 9 } to list out the remaining possibilities for round 3 and 4 groupings. Casework shows that there are 2 ways to group the players in the remaining two rounds. We multiply 2 280 · (3!) · 2 to get 20160.