HMMT 二月 2010 · 冲刺赛 · 第 28 题

英文解析

28. [ 18 ] Danielle Bellatrix Robinson is organizing a poker tournament with 9 people. The tournament willhave 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 roundgroups 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 }
    into each group. There are (3!) ways to do this. Now, because of symmetry, we can use the round 22
    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
    280 · (3!) · 2 to get 20160.2