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Difference between revisions of "2021 AMC 10A Problems/Problem 20"
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==Solution (bashing)==
 
==Solution (bashing)==
−
We write out the 120 cases.  
+
We write out the <math>120</math> cases.  
 
These cases are the ones that work:
 
These cases are the ones that work:
−
<math>1,3,2,5,4</math> \\
+
<math>1,3,2,5,4 \linebreak</math>
−
<math>1,4,2,5,3</math> \\
+
<math>1,4,2,5,3 \linebreak</math>
−
<math>1,4,3,5,2</math> \\
+
<math>1,4,3,5,2 \linebreak</math>
−
<math>1,5,2,4,3</math> \\
+
<math>1,5,2,4,3 \linebreak</math>
−
<math>1,5,3,4,2</math> \\
+
<math>1,5,3,4,2 \linebreak</math>
−
<math>2,1,4,3,5</math> \\
+
<math>2,1,4,3,5 \linebreak</math>
−
<math>2,1,5,3,4</math> \\
+
<math>2,1,5,3,4 \linebreak</math>
−
<math>2,3,1,5,4</math> \\
+
<math>2,3,1,5,4 \linebreak</math>
−
<math>2,4,1,5,3</math> \\
+
<math>2,4,1,5,3 \linebreak</math>
−
<math>2,4,3,5,1</math> \\
+
<math>2,4,3,5,1</math>
 
<math>2,5,1,4,3</math>
 
<math>2,5,1,4,3</math>
 
<math>2,5,3,4,1</math>
 
<math>2,5,3,4,1</math>

Revision as of 15:22, 11 February 2021

Problem

In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing? $\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44$

Solution (bashing)

We write out the $120$ cases. These cases are the ones that work: $1,3,2,5,4 \linebreak$ $1,4,2,5,3 \linebreak$ $1,4,3,5,2 \linebreak$ $1,5,2,4,3 \linebreak$ $1,5,3,4,2 \linebreak$ $2,1,4,3,5 \linebreak$ $2,1,5,3,4 \linebreak$ $2,3,1,5,4 \linebreak$ $2,4,1,5,3 \linebreak$ $2,4,3,5,1$ $2,5,1,4,3$ $2,5,3,4,1$ $3,1,4,2,5$ $3,1,5,2,4$ $3,2,4,1,5$ $3,2,5,1,4$ $3,4,1,5,2$ $3,4,2,5,1$ $3,5,1,4,2$ $3,5,2,4,1$ $4,1,3,2,5$ $4,1,5,2,3$ $4,2,3,1,5$ $4,2,5,1,3$ $4,3,5,1,2$ $4,5,1,3,2$ $4,5,2,3,1$ $5,1,3,2,4$ $5,1,4,2,3$ $5,2,3,1,4$ $5,2,4,1,3$ $5,3,4,1,2$ We count these out and get $\boxed{\text{D: }32}$ permutations that work. ~contactbibliophile

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