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Difference between revisions of "2001 AIME I Problems/Problem 14"
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==Solution 2 ( Less recursion than solution 1) ==
 
==Solution 2 ( Less recursion than solution 1) ==
−
Let <math>M_n</math> represent the number of mail delivery patterns that end with the last house receiving mail. This is <math>b_n</math> in Solution 1. Similarly define <math>A_n</math> to be the number of mail delivery patterns that end with last house not receiving mail. Let <math>T_n</math> be the total number of mail delivery patterns.
+
Let <math>M_n</math> represent the number of mail delivery patterns that end with the last house receiving mail. This is <math>b_n</math> in Solution 1. Similarly define <math>A_n</math> to be the number of mail delivery patterns that end with last house not receiving mail. This is just <math>a_n</math> and <math>c_n</math> in solution 1. Let <math>T_n</math> be the total number of mail delivery patterns.
  
 
Here are the possible ending cases: the string ends in <math>1, 10,</math> or <math>100</math>. The first case is just <math>M_n</math>. The second case is <math>M_{n-1}</math>. The third case is <math>M_{n-2}</math>. So we have <math>T_n = M_n + M_{n-1} + M_{n-2}</math>. Since we want <math>T_{19}</math>, it is just <math> M_{18} + M_{17} + M_{16}</math>.
 
Here are the possible ending cases: the string ends in <math>1, 10,</math> or <math>100</math>. The first case is just <math>M_n</math>. The second case is <math>M_{n-1}</math>. The third case is <math>M_{n-2}</math>. So we have <math>T_n = M_n + M_{n-1} + M_{n-2}</math>. Since we want <math>T_{19}</math>, it is just <math> M_{18} + M_{17} + M_{16}</math>.
−
Now using the same logic as above we can find <math>M_n = M_{n-2} + M{n-3}</math> ( the cases are 01 and 001). We can refer back to solution 1's table and only keep track of <math>b_n</math>, ignoring both <math>a_n</math> and <math>c_n</math>. - MathLegend27
+
Now using the same logic as above we can find <math>M_n = M_{n-2} + M_{n-3}</math> ( the cases are 01 and 001). We can refer back to solution 1's table and only keep track of <math>b_n</math>, ignoring both <math>a_n</math> and <math>c_n</math>.  
 +
- MathLegend27
  
 
== See also ==
 
== See also ==

Revision as of 19:08, 8 November 2018

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?

Solutions

Solution 1

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: \[\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \multicolumn{19}{c}{}\\\hline n&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\\\hline a_n&1&1&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86\\\hline b_n&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86&114&151\\\hline c_n&1&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86&114\\\hline \end{array}\]

As a result $a_{19}+b_{19}+c_{19}=\boxed{351}$.

Solution 2 ( Less recursion than solution 1)

Let $M_n$ represent the number of mail delivery patterns that end with the last house receiving mail. This is $b_n$ in Solution 1. Similarly define $A_n$ to be the number of mail delivery patterns that end with last house not receiving mail. This is just $a_n$ and $c_n$ in solution 1. Let $T_n$ be the total number of mail delivery patterns.

Here are the possible ending cases: the string ends in $1, 10,$ or $100$. The first case is just $M_n$. The second case is $M_{n-1}$. The third case is $M_{n-2}$. So we have $T_n = M_n + M_{n-1} + M_{n-2}$. Since we want $T_{19}$, it is just $M_{18} + M_{17} + M_{16}$. Now using the same logic as above we can find $M_n = M_{n-2} + M_{n-3}$ ( the cases are 01 and 001). We can refer back to solution 1's table and only keep track of $b_n$, ignoring both $a_n$ and $c_n$. - MathLegend27

See also

2001 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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