Difference between revisions of "2008 AMC 10B Problems/Problem 22"
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There is a block of RWRWR. The B can go in any slot between two letters. 6 ways. | There is a block of RWRWR. The B can go in any slot between two letters. 6 ways. | ||
− | =Case 2= | + | ===Case 2=== |
There is a RBR. Then, there is a WRW, and there are 2 orderings. | There is a RBR. Then, there is a WRW, and there are 2 orderings. | ||
+ | |||
=Case 3= | =Case 3= | ||
There is a WBW. Then, there is a RWR, and there are 2 orderings. | There is a WBW. Then, there is a RWR, and there are 2 orderings. |
Revision as of 11:26, 15 September 2019
Contents
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 .
Solution 3
There are 3 cases:
Case 1
There is a block of RWRWR. The B can go in any slot between two letters. 6 ways.
Case 2
There is a RBR. Then, there is a WRW, and there are 2 orderings.
Case 3
There is a WBW. Then, there is a RWR, and there are 2 orderings.
There are 60 ways total, and 10 valid ones, giving as the answer.
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.