2008 AMC 10B Problems/Problem 22

Problem

Three red beads, two white beads, and one blue bead are placed in line in random order. What is the probability that no two neighboring beads are the same color?

Solution 1

There are two ways to arrange the red beads, where represents a red bead and represents a blank space.

In the first, there are three ways to place a bead in the first free space, two for the second free space, and one for the third, so there are arrangements. In the second, a white bead must be placed in the third free space, so there are two possibilities for the third space, two for the second, and one for the first. That makes arrangements. There are arrangements in total. The two cases above can be reversed, so we double to arrangements. Also, in each case, there are three ways to place the first red bead, two for the second, and one for the third, so we multiply by to get arrangements. There are total arrangements so the answer is .

Solution 2

If you list the case for the red balls and blanks, these are four scenarios:

In the cases one and two, the white balls must go in the blank surrounded on either side by the red balls. Like this: However, now with only one white ball and one blue ball left, there are two ways to order them in both cases. So for a total of possibilities.

In the last two cases, the two white balls and the blue ball can go anywhere, in those three blanks, becuase they are separated by a red ball. So there are ways for each case. This adds up to a total of possibilities for the last two cases.

Adding them up, we get total orderings.

There are total orderings.

So the answer is .

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.