HMMT 二月 2007 · COMB 赛 · 第 6 题

HMMT February 2007 — COMB Round — Problem 6

[ 5 ] Kevin has four red marbles and eight blue marbles. He arranges these twelve marbles randomly, in a ring. Determine the probability that no two red marbles are adjacent.

解析

[ 5 ] Kevin has four red marbles and eight blue marbles. He arranges these twelve marbles randomly, in a ring. Determine the probability that no two red marbles are adjacent. 7 Answer: . Select any blue marble and consider the remaining eleven marbles, arranged in a line. 33 The proportion of arrangement for which no two red marbles are adjacent will be the same as for the original twelve marbles, arranged in a ring. The total number of ways of arranging 4 red marbles out ( ) 11 of 11 is = 330. To count the number of arrangements such that no two red marbles are adjacent, 4 there must be one red marble between each two would-be adjacent red marbles. Having fixed the positions of three blue marbles we have four blue marbles to play with. So that we can arrange the ( ) 8 remaining four marbles is = 70 ways. This yields a probability of 70 / 330 = 7 / 33 as our final 4 answer.