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Difference between revisions of "2003 AIME II Problems/Problem 3"
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== Problem ==
 
== Problem ==
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Define a <math>\text{good~word}</math> as a sequence of letters that consists only of the letters <math>A</math>, <math>B</math>, and <math>C</math> - some of these letters may not appear in the sequence - and in which <math>A</math> is never immediately followed by <math>B</math>, <math>B</math> is never immediately followed by <math>C</math>, and <math>C</math> is never immediately followed by <math>A</math>. How many seven-letter good words are there?
  
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== Solution ==
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== Solution 1 ==  
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{{solution}}
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There are three letters to make the first letter in the sequence. However, after the first letter (whatever it is), only two letters can follow it, since one of the letters is restricted. Therefore, the number of seven-letter good words is <math>3*2^6=192</math>
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Therefore, there are <math>\boxed{192}</math> seven-letter good words.
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== Solution 2 ==
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There are three choices for the first letter and two choices for each subsequent letter, so there are <math>3\cdot2^{n-1}\ n</math>-letter good words. Substitute <math>n=7</math> to find there are <math>3\cdot2^6=\boxed{192}</math> seven-letter good words. ~ aopsav (Credit to AoPS Alcumus)
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== Solution 3 (Recursion) ==
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We solve this problem using recursion. Let <math>f(x)</math> be the number of <math>x</math>-letter good words. Thus <math>f(1) = 3</math> (A, B or C) and the answer is just <math>f(7)</math>.  The recurrence relation can be found by considering the last letter of one of the valid strings of length <math>x - 1</math>.  There are <math>2</math> possibilities for the next letter and thus <math>f(x) = 2 \cdot f(x-1)</math>.  Now we can find a closed form as <math>f(x) = 3 \cdot 2 ^{x-1}</math>  (easy to prove by induction) and thus <math>f(7) = 64 * 3 = \boxed{192}</math> seven-letter good words.  ~AK2006
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== Video Solution by Sal Khan ==
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https://www.youtube.com/watch?v=JU67TL2L1CA&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=9
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- AMBRIGGS
  
 
== See also ==
 
== See also ==
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* [[2003 AIME II Problems/Problem 2| Previous problem]]
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{{AIME box|year=2003|n=II|num-b=2|num-a=4}}
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* [[2003 AIME II Problems/Problem 4| Next problem]]
 
  
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* [[2003 AIME II Problems]]
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[[Category: Intermediate Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 16:29, 30 July 2022

Problem

Define a $\text{good~word}$ as a sequence of letters that consists only of the letters $A$, $B$, and $C$ - some of these letters may not appear in the sequence - and in which $A$ is never immediately followed by $B$, $B$ is never immediately followed by $C$, and $C$ is never immediately followed by $A$. How many seven-letter good words are there?

Solution 1

There are three letters to make the first letter in the sequence. However, after the first letter (whatever it is), only two letters can follow it, since one of the letters is restricted. Therefore, the number of seven-letter good words is $3*2^6=192$

Therefore, there are $\boxed{192}$ seven-letter good words.

Solution 2

There are three choices for the first letter and two choices for each subsequent letter, so there are $3\cdot2^{n-1}\ n$-letter good words. Substitute $n=7$ to find there are $3\cdot2^6=\boxed{192}$ seven-letter good words. ~ aopsav (Credit to AoPS Alcumus)

Solution 3 (Recursion)

We solve this problem using recursion. Let $f(x)$ be the number of $x$-letter good words. Thus $f(1) = 3$ (A, B or C) and the answer is just $f(7)$. The recurrence relation can be found by considering the last letter of one of the valid strings of length $x - 1$. There are $2$ possibilities for the next letter and thus $f(x) = 2 \cdot f(x-1)$. Now we can find a closed form as $f(x) = 3 \cdot 2 ^{x-1}$ (easy to prove by induction) and thus $f(7) = 64 * 3 = \boxed{192}$ seven-letter good words. ~AK2006

Video Solution by Sal Khan

https://www.youtube.com/watch?v=JU67TL2L1CA&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=9 - AMBRIGGS

See also

2003 AIME II (Problems • Answer Key • Resources)
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Problem 4
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All AIME Problems and Solutions

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