Difference between revisions of "2008 AMC 10B Problems/Problem 22"
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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? | 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? | ||
| − | A) 1/12 B) 1/10 C) 1/6 D) 1/3 E) 1/2 | + | <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== | ==Solution== | ||
There are two ways to arrange the red beads. | There are two ways to arrange the red beads. | ||
| − | + | # <tt>R _ R _ R _</tt> | |
| − | + | # <tt>R _ _ R _ R</tt> | |
| − | In the first, there are three ways to place a bead in the first space, two for the second space, and one for the third, so there are 6 arrangements. | + | 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 <math>6</math> arrangements. |
| − | In the second, a white bead must be placed in the third space, so there are two possibilities for the third space, two for the second, and one for the first. That makes 4 arrangements. There are 6+4=10 arrangements in total. The two cases above can be reversed, so we double 10 to 20 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 6 to get 20 | + | 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 <math>4</math> arrangements. There are <math>6+4=10</math> arrangements in total. The two cases above can be reversed, so we double <math>10</math> to <math>20</math> 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 <math>6</math> to get <math>20\cdot 6 = 120</math> arrangements. There are <math>6! = 720</math> total arrangements so the answer is <math>120/720 = \boxed{1/6}</math>. |
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=21|num-a=23}} | {{AMC10 box|year=2008|ab=B|num-b=21|num-a=23}} | ||
Revision as of 15:11, 25 January 2009
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
There are two ways to arrange the red beads.
- R _ R _ R _
- R _ _ R _ R
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 
.
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 | ||
