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Difference between revisions of "2008 AMC 10B Problems/Problem 22"
(→‎Solution 2 (Order Doesn't Matter))
 
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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==
+
==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> - </math> represents a blank space.
  
−
<math>1. R B R B R B </math>
+
<math>1. R - R - R - </math>
  
−
<math>2. R B B R B R </math>
+
<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.
−
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 get <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==
+
==Solution 2 (Order Doesn't Matter)==
  
−
If you list the case for the red balls and blanks, these are four scenarios:
+
If you list the cases for the red balls and blanks, these are four scenarios:
  
 
<math> 1. R - - R - R </math>
 
<math> 1. R - - R - R </math>
Line 26: Line 26:
 
<math> 4. - 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.  
+
In cases 1 and 2, the white balls must go in the blank surrounded on either side by the red balls.  
 
Like this:
 
Like this:
 
<math> 1. R - - R W R </math>
 
<math> 1. R - - R W R </math>
Line 32: Line 32:
 
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.
 
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.
+
In the last two cases, the two white balls and the blue ball can go anywhere in those three blanks, because 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.
 
Adding them up, we get <math>10</math> total orderings.
Line 43: Line 43:
 
There are 3 cases:
 
There are 3 cases:
 
===Case 1===
 
===Case 1===
−
There is a block of RWRWR. The B can go in any slot between two letters. 6 ways.
+
There is a block of <math>RWRWR</math>. The <math>B</math> 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 <math>RBR</math>. Then, there is a <math>WRW</math>, 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 <math>WBW</math>. Then, there is a <math>RWR</math>, and there are 2 orderings.
  
 
There are 60 ways total, and 10 valid ones, giving <math>\boxed{\frac{1}{6}}</math> as the answer.
 
There are 60 ways total, and 10 valid ones, giving <math>\boxed{\frac{1}{6}}</math> as the answer.
Line 68: Line 68:
 
<math>W R\ W R\ R</math>
 
<math>W R\ W R\ R</math>
  
−
For the first case, there are <math>6</math> 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 <math>\frac{6+4(1)}{60}=\frac{10}{60}=\boxed{\mathrm{(A) }\frac16}</math>
+
For the first case, there are <math>6</math> 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 <math>\frac{6+4(1)}{60}=\frac{10}{60}=\boxed{\mathrm{(C) }\frac16}</math>
 +
 
 +
==Solution 5 (Video Solution)==
 +
Video: https://youtu.be/qnAAzoRANG4
  
 
==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}}

Latest revision as of 12:49, 7 June 2021

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 1 (Order Matters)

There are two ways to arrange the red beads, where $R$ represents a red bead and $-$ represents a blank space.

$1. R - R - R -$

$2. R - - R - R$

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 get $20$ 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 $120/720 = \boxed{1/6}$.

Solution 2 (Order Doesn't Matter)

If you list the cases for the red balls and blanks, these are four scenarios:

$1. R - - R - R$

$2. R - R - - R$

$3. R - R - R -$

$4. - R - R - R$

In cases 1 and 2, the white balls must go in the blank surrounded on either side by the red balls. Like this: $1. R - - R W R$ $2. R W R - - R$ 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 $2 + 2 = 4$ possibilities.

In the last two cases, the two white balls and the blue ball can go anywhere in those three blanks, because they are separated by a red ball. So there are $3$ ways for each case. This adds up to a total of $6$ possibilities for the last two cases.

Adding them up, we get $10$ total orderings.

There are $\frac{6!}{3! * 2! * 1!} = 60$ total orderings.

So the answer is $\frac{10}{60} = \boxed{1/6}$.

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 $\boxed{\frac{1}{6}}$ as the answer.

Solution 4

There are $\frac{6!}{3!\cdot2!\cdot1!}=60$ 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:

$R\ W R\ W R$

$R\ R\ W R\ W$

$W R\ R\ W R$

$R\ W R\ R\ W$

$W R\ W R\ R$

For the first case, there are $6$ 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 $\frac{6+4(1)}{60}=\frac{10}{60}=\boxed{\mathrm{(C) }\frac16}$

Solution 5 (Video Solution)

Video: https://youtu.be/qnAAzoRANG4

See also

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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All AMC 10 Problems and Solutions

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