2015 AIME II Problems/Problem 12
Problem
There are possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.
Solution
The solution is a simple recursion:
We have three cases for the ending of a string: three in a row, two in a row, and a single:
...AAA ...BAA ...BBA
For case , we could only add a B to the end, making it a case . For case , we could add an A or a B to the end, making it a case if you add an A, or a case if you add a B. For case , we could add an A or a B to the end, making it a case or a case .
Let us create three series to represent the number of permutations for each case: , , and representing case , , and respectively.
The series have the following relationship:
For : and both equal , . With some simple math, we have: , , and . Summing the three up we have our solution: .
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.