Difference between revisions of "2008 AMC 10B Problems/Problem 22"

(Solution)
Line 7: Line 7:
 
There are two ways to arrange the red beads.
 
There are two ways to arrange the red beads.
  
# <tt>R _ R _ R _</tt>
+
<math>1. R - R -R -</math>
# <tt>R _ _ R _ R</tt>
+
 
 +
<math>2. R -- R - R</math>
  
 
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 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.

Revision as of 00:10, 1 February 2015

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?

$\mathrm{(A)}\ 1/12\qquad\mathrm{(B)}\ 1/10\qquad\mathrm{(C)}\ 1/6\qquad\mathrm{(D)}\ 1/3\qquad\mathrm{(E)}\ 1/2$

Solution

There are two ways to arrange the red beads.

$1. R - R -R -$

$2. R -- R - R$

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 $6$ 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 $4$ 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 $20\cdot 6 = 120$ arrangements. There are $6! = 720$ total arrangements so the answer is $120/720 = \boxed{1/6}$.

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
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

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