Difference between revisions of "2008 AIME I Problems/Problem 11"
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</cmath> | </cmath> | ||
By counting, we find that <math>a_1 = 0, b_1 = 1, a_2 = 1, b_2 = 0</math>. | By counting, we find that <math>a_1 = 0, b_1 = 1, a_2 = 1, b_2 = 0</math>. | ||
− | <cmath>\begin{ | + | <cmath>\begin{array}{|r||r|r|||r||r|r|} |
\hline | \hline | ||
n & a_n & b_n & n & a_n & b_n\ | n & a_n & b_n & n & a_n & b_n\ | ||
Line 33: | Line 33: | ||
14&80&92\ | 14&80&92\ | ||
\hline | \hline | ||
− | \end{ | + | \end{array} |
</cmath> | </cmath> | ||
Therefore, the number of such strings of length <math>14</math> is <math>a_{14} + b_{14} = \boxed{172}</math>. | Therefore, the number of such strings of length <math>14</math> is <math>a_{14} + b_{14} = \boxed{172}</math>. | ||
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We replace "14" with "<math>2n</math>". We first note that we must have an even number of chunks of <math>B</math>'s, because of parity issues. We then note that every chunk of <math>B</math>'s except the last one must end in the sequence <math>BAA</math>, since we need at least two <math>A</math>'s to separate it from the next chunk of <math>B</math>'s. The last chunk of <math>B</math>'s must, of course, end with a <math>B</math>. Thus our sequence must look like this : | We replace "14" with "<math>2n</math>". We first note that we must have an even number of chunks of <math>B</math>'s, because of parity issues. We then note that every chunk of <math>B</math>'s except the last one must end in the sequence <math>BAA</math>, since we need at least two <math>A</math>'s to separate it from the next chunk of <math>B</math>'s. The last chunk of <math>B</math>'s must, of course, end with a <math>B</math>. Thus our sequence must look like this : | ||
<cmath> | <cmath> | ||
− | \boxed{\quad A\text{'s} \quad | + | \boxed{\quad A\text{'s} \quad} \boxed{\quad B\text{'s} \quad} BAA \boxed{\quad A\text{'s} \quad} \cdots \boxed{\quad B\text{'s} \quad}BAA \boxed{\quad A\text{'s} \quad} \boxed{\quad B\text{'s} \quad} B \boxed{\quad A\text{'s} \quad} , |
</cmath> | </cmath> | ||
where each box holds an even number of letters (possibly zero). | where each box holds an even number of letters (possibly zero). |
Revision as of 16:30, 13 March 2015
Problem
Consider sequences that consist entirely of 's and 's and that have the property that every run of consecutive 's has even length, and every run of consecutive 's has odd length. Examples of such sequences are , , and , while is not such a sequence. How many such sequences have length 14?
Contents
[hide]Solution
Solution 1
Let and denote, respectively, the number of sequences of length ending in and . If a sequence ends in an , then it must have been formed by appending two s to the end of a string of length . If a sequence ends in a it must have either been formed by appending one to a string of length ending in an , or by appending two s to a string of length ending in a . Thus, we have the recursions By counting, we find that . Therefore, the number of such strings of length is .
Solution 2
We replace "14" with "". We first note that we must have an even number of chunks of 's, because of parity issues. We then note that every chunk of 's except the last one must end in the sequence , since we need at least two 's to separate it from the next chunk of 's. The last chunk of 's must, of course, end with a . Thus our sequence must look like this : where each box holds an even number of letters (possibly zero).
If we want a sequence with chunks of 's, then we have letters with which to fill the boxes. Since each box must have an even number of letters, we may put the letters in the boxes in pairs. Then we have pairs of letters to put into boxes. By a classic balls-and-bins argument, the number of ways to do this is It follows that the total number of desirable sequences is For , this evaluates as .
See also
2008 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.