Difference between revisions of "2003 AIME II Problems/Problem 3"

Revision as of 01:56, 7 October 2020

Problem

Define a $good~word$ 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 * f(x-1). Now we can find a closed form as f(x) = 3 * 2 ^ (x - 1) (easy to prove by induction) and thus f(7) = 64 * 3 = $\boxed{192}$ seven-letter good words. ~AK2006

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 == 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 * f(x-1).  Now we can find a closed form as f(x) = 3 * 2 ^ (x - 1)  (easy to prove by induction) and thus f(7) = 64 * 3 = <math>\boxed{192}</math> seven-letter good words.
+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 * f(x-1).  Now we can find a closed form as f(x) = 3 * 2 ^ (x - 1)  (easy to prove by induction) and thus f(7) = 64 * 3 = <math>\boxed{192}</math> seven-letter good words.  ~AK2006
 == See also ==

2003 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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