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2001 AIME I Problems/Problem 14

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Problem

A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?

Solution

Let 0 represent a house that does not receive mail and 1 represent a house that does receive mail. This problem is now asking for the number of 19-digit strings of 0's and 1's such that there are no two consecutive 1's and no three consecutive 0's.

The last two digits of any n-digit string can't be 11, so the only possibilities are 00, 01, and 10.

Let a_n be the number of n-digit strings ending in 00, b_n be the number of n-digit strings ending in 01, and c_n be the number of n-digit strings ending in 10.

If an n-digit string ends in 00, then the previous digit must be a 1, and the last two digits of the n-1 digits substring will be 10. So a_{n} = c_{n-1}.

If an n-digit string ends in 01, then the previous digit can be either a 0 or a 1, and the last two digits of the n-1 digits substring can be either 00 or 10. So b_{n} = a_{n-1} + c_{n-1}.

If an n-digit string ends in 10, then the previous digit must be a 0, and the last two digits of the n-1 digits substring will be 01. So c_{n} = b_{n-1}.

Clearly, a_2=b_2=c_2=1. Using the recursive equations and initial values: \begin{tabular}[t]{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\multicolumn{19}{c}{}\\\hlinen&2&3&4&5&6&7...

Therefore, the number of 19-digit strings is a_{19}+b_{19}+c_{19} = 86+151+114 = \boxed{351}.

See also

2001 AIME I (ProblemsResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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