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> | ||
| − | ==Solution== | + | ==Solution 1== |
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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 started on the left side. Instead of starting at the C, let's start at the T at the end and work backwards. There is 1 option to choose the T. Then, there are 2 options to choose the S, 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 ways, so the answer is (E) none of these. | ||
Revision as of 23:10, 1 January 2024
Problem
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?
Solution 1
The path will either start on the left side, start on the right side, or go down the middle. Say it started on the left side. Instead of starting at the C, let's start at the T at the end and work backwards. There is 1 option to choose the T. Then, there are 2 options to choose the S, 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 ways, so the answer is (E) none of these.
