Difference between revisions of "2008 AMC 10B Problems/Problem 22"
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<math>\mathrm{(A)}\ 1/12\qquad\mathrm{(B)}\ 1/10\qquad\mathrm{(C)}\ 1/6\qquad\mathrm{(D)}\ 1/3\qquad\mathrm{(E)}\ 1/2</math> | <math>\mathrm{(A)}\ 1/12\qquad\mathrm{(B)}\ 1/10\qquad\mathrm{(C)}\ 1/6\qquad\mathrm{(D)}\ 1/3\qquad\mathrm{(E)}\ 1/2</math> | ||
| − | ==Solution 1(Order Matters)== | + | ==Solution 1 (Order Matters)== |
There are two ways to arrange the red beads, where <math> R </math> represents a red bead and <math> B </math> represents a blank space. | There are two ways to arrange the red beads, where <math> R </math> represents a red bead and <math> B </math> represents a blank space. | ||
Revision as of 07:44, 8 December 2020
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 (Order Matters)
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 (Order Doesn't Matter)
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.
Solution 4
There are 
total orderings.
Suppose we order the red and white beads first. If these two colors are ordered first, we must make sure that no neighboring beads are the same color, or only one pair of neighboring beads are the same color. There are five possible orderings:
For the first case, there are possible places we can put the blue bead. For the other 4 cases, there is only one place we can put the blue bead, which is between the pair of red beads. The desired probability is 
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. 
