HMMT 十一月 2012 · 冲刺赛 · 第 24 题

HMMT November 2012 — Guts Round — Problem 24

[ 12 ] 12 children sit around a circle. Find the number of ways we can distribute 4 pieces of candy among them such that each child gets at most one piece of candy and no two adjacent children both get a piece of candy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND 2 2 2 sin B + sin C − sin A

解析

[ 12 ] Answer: 105 Since 4 pieces of candy are distributed, there must be exactly 8 children who do not receive any candy; since no two consecutive children do receive candy, the 8 who do not must consist of 4 groups of consecutive children. We divide into cases based on the sizes of these groups: • { 5 , 1 , 1 , 1 } : there are 12 places to begin the group of 5 children who do not receive any candy • { 4 , 2 , 1 , 1 } : there are 12 places to begin the group of 4 children who do not receive candy and then 3 choices for the group of 2 children which does not receive candy, for a total of 36 choices • { 3 , 3 , 1 , 1 } : these 8 children can either be bunched in the order 3,3,1,1, or in the order 3,1,3,1; the first has 12 positions in which to begin the first group of 3 non-candy receiving children and the second has 6 possibilities (due to symmetry), for a total of 18 • { 3 , 2 , 2 , 1 } : there are 12 places to begin the group of 3 children who do not receive candy and then 3 choices for the group of 1 child which does not receive candy, for a total of 36 choices • { 2 , 2 , 2 , 2 } : there are 12 / 4 = 3 ways in which this can occur This gives a total of 12 + 36 + 18 + 36 + 3 = 105