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Difference between revisions of "1977 AHSME Problems/Problem 20"

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<math>\textbf{(A) }63\qquad \textbf{(B) }128\qquad \textbf{(C) }129\qquad \textbf{(D) }255\qquad  \textbf{(E) }\text{none of these}</math>
 
<math>\textbf{(A) }63\qquad \textbf{(B) }128\qquad \textbf{(C) }129\qquad \textbf{(D) }255\qquad  \textbf{(E) }\text{none of these}</math>
  
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==Solution 1==
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==Solution==
  
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The path will either start on the left side, start on the right side, or go down the middle. Say it starts on the left side. Instead of counting it starting from the <math>C</math>, let's start at the T and work backwards. Starting at the <math>T</math>, there are <math>2</math> options to choose the <math>S</math>. Then, there are <math>2</math> options to chose the <math>E</math>, <math>T</math>, <math>N</math>, <math>O</math>, and <math>C</math>, for a total of <math>2^6=64</math> possibilities. But we need to subtract the case where it does down the middle, which makes it <math>63</math> possibilities. The right side also has <math>63</math> possibilities by symmetry. There is <math>1</math> way for it to go down the middle, for a total of <math>63+63+1=127</math> ways, so the answer is <math>\textbf{(E) }\text{none of these}</math>.
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The path will either start on the left side, start on the right side, or go down the middle. Say it starts on the left side. Instead of counting it starting from the <math>C</math>, let's start at the T and work backwards. Starting at the <math>T</math>, there are <math>2</math> options to choose the <math>S</math>. Then, there are <math>2</math> options to chose the <math>E</math>, <math>T</math>, <math>N</math>, <math>O</math>, and <math>C</math>, for a total of <math>2^6=64</math> possibilities. But we need to subtract the case where it does down the middle, which makes it <math>63</math> possibilities. The right side also has <math>63</math> possibilities by symmetry. There is <math>1</math> way for it to go down the middle, for a total of <math>63+63+1=127</math> paths, so the answer is <math>\textbf{(E) }\text{none of these}</math>.
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~alexanderruan

Latest revision as of 23:25, 1 January 2024

Problem

$\begin{tabular}{ccccccccccccc}& & & & & & C & & & & & &\\ & & & & & C & O & C & & & & &\\ & & & & C & O & N & O & C & & & &\\ & & & C & O & N & T & N & O & C & & &\\ & & C & O & N & T & E & T & N & O & C & &\\ & C & O & N & T & E & S & E & T & N & O & C &\\ C & O & N & T & E & S & T & S & E & T & N & O & C \end{tabular}$


For how many paths consisting of a sequence of horizontal and/or vertical line segments, with each segment connecting a pair of adjacent letters in the diagram above, is the word CONTEST spelled out as the path is traversed from beginning to end?

$\textbf{(A) }63\qquad \textbf{(B) }128\qquad \textbf{(C) }129\qquad \textbf{(D) }255\qquad \textbf{(E) }\text{none of these}$

Solution

The path will either start on the left side, start on the right side, or go down the middle. Say it starts on the left side. Instead of counting it starting from the $C$, let's start at the T and work backwards. Starting at the $T$, there are $2$ options to choose the $S$. Then, there are $2$ options to chose the $E$, $T$, $N$, $O$, and $C$, for a total of $2^6=64$ possibilities. But we need to subtract the case where it does down the middle, which makes it $63$ possibilities. The right side also has $63$ possibilities by symmetry. There is $1$ way for it to go down the middle, for a total of $63+63+1=127$ paths, so the answer is $\textbf{(E) }\text{none of these}$.

~alexanderruan

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