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== | + | ==Solution 1== |
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. | ||
| Line 13: | Line 13: | ||
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. | ||
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>. | 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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | If you list the case for the red balls and blanks, these are four scenarios: | ||
| + | |||
| + | <math>1. R - - R - R </math> | ||
| + | <math>2. R - R - - R </math> | ||
| + | <math>3. R - R - R - </math> | ||
| + | <math>4. - R - R - R </math> | ||
| + | |||
| + | In the cases one and two, the white balls must go in the blank surrounded on either side by the red balls. | ||
| + | Like this: | ||
| + | <math> 1. R - - R W R </math> | ||
| + | <math> 2. R W R - - R </math> | ||
| + | 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 <math>2 + 2 = 4</math> 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 <math>3</math> ways for each case. This adds up to a total of <math>6</math> possibilities for the last two cases. | ||
| + | |||
| + | Adding them up, we get <math>10</math> total orderings. | ||
| + | |||
| + | There are <math>\frac{6!}{3! * 2! * 1!} = 60</math> total orderings. | ||
| + | |||
| + | So the answer is <math>\frac{10}{60} = \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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:56, 31 January 2018
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 
.
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. 
