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Difference between revisions of "2019 AMC 10B Problems/Problem 25"

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This is because for any valid sequence of length <math>n</math>, you can append either <math>10</math> or <math>110</math> and the resulting sequence will still satisfy the given conditions.
 
This is because for any valid sequence of length <math>n</math>, you can append either <math>10</math> or <math>110</math> and the resulting sequence will still satisfy the given conditions.
  
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It is easy to find <math>f(5) = 1</math> and <math>f(6) = 2</math> by hand, and then by the recursive formula, we have <math>f(19) = \boxed{\textbf{(C) }65}</math>.
+
It is easy to find <math>f(5) = 1</math> with the only possible sequence being <math>01010</math> and <math>f(6) = 2</math> with the only two possible sequences being <math>011010</math> and <math>010110</math> by hand, and then by the recursive formula, we have <math>f(19) = \boxed{\textbf{(C) }65}</math>.
  
 
==Solution 2 (casework)==
 
==Solution 2 (casework)==
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'''Case 2''': two <math>3</math>s and six <math>2</math>s - there are <math>{8\choose2} = 28</math> ways to arrange them.
 
'''Case 2''': two <math>3</math>s and six <math>2</math>s - there are <math>{8\choose2} = 28</math> ways to arrange them.
  
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'''Case 3''': four <math>3</math>s and three <math>2</math>s - there are <math>{7\choose3} = 35</math> ways to arrange them.
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'''Case 3''': four <math>3</math>s and three <math>2</math>s - there are <math>{7\choose4} = 35</math> ways to arrange them.
  
 
'''Case 4''': six <math>3</math>s - there is only <math>1</math> way to arrange them.
 
'''Case 4''': six <math>3</math>s - there is only <math>1</math> way to arrange them.
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Summing the four cases gives <math>1+28+35+1=\boxed{\textbf{(C) }65}</math>.
 
Summing the four cases gives <math>1+28+35+1=\boxed{\textbf{(C) }65}</math>.
  
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==Solution 3(Simpler)==
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==Solution 3 (casework and blocks)==
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Firstly, notice how the ending of a suitable string could be a 010, or a 0110. This information basically gives us the answer, as we know that for any string of length N, the possible solutions, would be a string of length n-2, with 10 appended, and a string of length n-3, with 110 appended.  
+
We can simplify the original problem into a problem where there are <math>2^{17}</math> binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of <math>0</math>s, <math>1</math>s, and <math>11</math>s. Now, we use casework:
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For example:
 
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The strings of length 4 are 0110, the strings of length 5 are 01010, so to make length 7 we can add a 10 to length 5, 0101010, and then add 110 to the string of length 4, 0110110, these are the only possible strings, since the endings have to be 10, or 110, and the strings of length 5 and 4, work and can be modified to work as a 7, since they end with 0.
 
  
 +
'''Case 1''': Alternating 1s and 0s. There is simply 1 way to do this: <math>0101010101010101010</math>.
 +
Now, we note that there cannot be only one block of <math>11</math> in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of <math>11</math>s this cannot be satisfied. This is true for all odd numbers of <math>11</math> blocks.
  
 +
'''Case 2''': There are 2 <math>11</math> blocks. Using the zeroes in the sequence as dividers, we have a sample as <math>0110110101010101010</math>. We know there are 8 places for <math>11</math>s, which will be filled by <math>1</math>s if the <math>11</math>s don't fill them. This is <math>{8\choose2} = 28</math> ways.
  
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Continue this process until 19 and you get  <math> \boxed{\textbf{(C) }65}</math>.
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'''Case 3''': Four <math>11</math> blocks arranged. Using the same logic as Case 2, we have <math>{7\choose4} = 35</math> ways to arrange four <math>11</math> blocks.
 +
 
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'''Case 4''': No single <math>1</math> blocks, only <math>11</math> blocks. There is simply one case for this, which is <math>0110110110110110110</math>.
 +
 
 +
Adding these four cases, we have <math>1+28+35+1=\boxed{\textbf{(C) }65}</math> as our final answer.  
 +
 
 +
~Equinox8
  
 
==Video Solution==
 
==Video Solution==
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{{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}
 
{{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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SUB2PEWDS
 

Revision as of 13:29, 25 June 2020

The following problem is from both the 2019 AMC 10B #25 and 2019 AMC 12B #23, so both problems redirect to this page.

Problem

How many sequences of $0$s and $1$s of length $19$ are there that begin with a $0$, end with a $0$, contain no two consecutive $0$s, and contain no three consecutive $1$s?

$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$

Solution 1 (recursion)

We can deduce, from the given restrictions, that any valid sequence of length $n$ will start with a $0$ followed by either $10$ or $110$. Thus we can define a recursive function $f(n) = f(n-3) + f(n-2)$, where $f(n)$ is the number of valid sequences of length $n$.

This is because for any valid sequence of length $n$, you can append either $10$ or $110$ and the resulting sequence will still satisfy the given conditions.

It is easy to find $f(5) = 1$ with the only possible sequence being $01010$ and $f(6) = 2$ with the only two possible sequences being $011010$ and $010110$ by hand, and then by the recursive formula, we have $f(19) = \boxed{\textbf{(C) }65}$.

Solution 2 (casework)

After any particular $0$, the next $0$ in the sequence must appear exactly $2$ or $3$ positions down the line. In this case, we start at position $1$ and end at position $19$, i.e. we move a total of $18$ positions down the line. Therefore, we must add a series of $2$s and $3$s to get $18$. There are a number of ways to do this:

Case 1: nine $2$s - there is only $1$ way to arrange them.

Case 2: two $3$s and six $2$s - there are ${8\choose2} = 28$ ways to arrange them.

Case 3: four $3$s and three $2$s - there are ${7\choose4} = 35$ ways to arrange them.

Case 4: six $3$s - there is only $1$ way to arrange them.

Summing the four cases gives $1+28+35+1=\boxed{\textbf{(C) }65}$.

Solution 3 (casework and blocks)

We can simplify the original problem into a problem where there are $2^{17}$ binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of $0$s, $1$s, and $11$s. Now, we use casework:

Case 1: Alternating 1s and 0s. There is simply 1 way to do this: $0101010101010101010$. Now, we note that there cannot be only one block of $11$ in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of $11$s this cannot be satisfied. This is true for all odd numbers of $11$ blocks.

Case 2: There are 2 $11$ blocks. Using the zeroes in the sequence as dividers, we have a sample as $0110110101010101010$. We know there are 8 places for $11$s, which will be filled by $1$s if the $11$s don't fill them. This is ${8\choose2} = 28$ ways.

Case 3: Four $11$ blocks arranged. Using the same logic as Case 2, we have ${7\choose4} = 35$ ways to arrange four $11$ blocks.

Case 4: No single $1$ blocks, only $11$ blocks. There is simply one case for this, which is $0110110110110110110$.

Adding these four cases, we have $1+28+35+1=\boxed{\textbf{(C) }65}$ as our final answer.

~Equinox8

Video Solution

For those who want a video solution: https://youtu.be/VamT49PjmdI

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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
2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 12 Problems and Solutions

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