Difference between revisions of "2019 AMC 10B Problems/Problem 25"
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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. | 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. | ||
| − | + | Continue this process until 19 and you get <math> \boxed{\textbf{(C) }65}</math>. | |
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==Video Solution== | ==Video Solution== | ||
Revision as of 09:21, 28 February 2019
- The following problem is from both the 2019 AMC 10B #25 and 2019 AMC 12B #23, so both problems redirect to this page.
Contents
Problem
How many sequences of 
s and 
s of length 
are there that begin with a , end with a , contain no two consecutive s, and contain no three consecutive s?
Solution 1 (recursion)
We can deduce, from the given restrictions, that any valid sequence of length 
will start with a followed by either 
or 
.
Thus we can define a recursive function 
, where 
is the number of valid sequences of length .
This is because for any valid sequence of length , you can append either or and the resulting sequence will still satisfy the given conditions.
It is easy to find 
and 
by hand, and then by the recursive formula, we have 
.
Solution 2 (casework)
After any particular , the next in the sequence must appear exactly 
or 
positions down the line. In this case, we start at position and end at position , i.e. we move a total of 
positions down the line. Therefore, we must add a series of s and s to get . There are a number of ways to do this:
Case 1: nine s - there is only way to arrange them.
Case 2: two s and six s - there are 
ways to arrange them.
Case 3: four s and three s - there are 
ways to arrange them.
Case 4: six s - there is only way to arrange them.
Summing the four cases gives 
.
Solution 3(Simpler)
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.
For example:
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.
Continue this process until 19 and you get 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
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