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Difference between revisions of "2021 AMC 10A Problems/Problem 20"

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The AMC 10 has not started yet.
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==Problem==
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In how many ways can the sequence <math>1,2,3,4,5</math> be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
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<math>\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44</math>
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 +
==Solution (bashing)==
 +
We write out the 120 cases.
 +
[hide=The Cases]
 +
<math>1,2,3,4,5</math>
 +
<math>1,2,3,5,4</math>
 +
<math>1,2,4,3,5</math>
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<math>1,2,4,5,3</math>
 +
<math>1,2,5,3,4</math>
 +
<math>1,2,5,4,3</math>
 +
<math>1,3,2,4,5</math>
 +
<math>1,3,2,5,4</math>
 +
<math>1,3,4,2,5</math>
 +
<math>1,3,4,5,2</math>
 +
<math>1,3,5,2,4</math>
 +
<math>1,3,5,4,2</math>
 +
<math>1,4,2,3,5</math>
 +
<math>1,4,2,5,3</math>
 +
<math>1,4,3,2,5</math>
 +
<math>1,4,3,5,2</math>
 +
<math>1,4,5,2,3</math>
 +
<math>1,4,5,3,2</math>
 +
<math>1,5,2,3,4</math>
 +
<math>1,5,2,4,3</math>
 +
<math>1,5,3,2,4</math>
 +
<math>1,5,3,4,2</math>
 +
<math>1,5,4,2,3</math>
 +
<math>1,5,4,3,2</math>
 +
<math>2,1,3,4,5</math>
 +
<math>2,1,3,5,4</math>
 +
<math>2,1,4,3,5</math>
 +
<math>2,1,4,5,3</math>
 +
<math>2,1,5,3,4</math>
 +
<math>2,1,5,4,3</math>
 +
<math>2,3,1,4,5</math>
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<math>2,3,1,5,4</math>
 +
<math>2,3,4,1,5</math>
 +
<math>2,3,4,5,1</math>
 +
<math>2,3,5,1,4</math>
 +
<math>2,3,5,4,1</math>
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<math>2,4,1,3,5</math>
 +
<math>2,4,1,5,3</math>
 +
<math>2,4,3,1,5</math>
 +
<math>2,4,3,5,1</math>
 +
<math>2,4,5,1,3</math>
 +
<math>2,4,5,3,1</math>
 +
<math>2,5,1,3,4</math>
 +
<math>2,5,1,4,3</math>
 +
<math>2,5,3,1,4</math>
 +
<math>2,5,3,4,1</math>
 +
<math>2,5,4,1,3</math>
 +
<math>2,5,4,3,1</math>
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<math>3,1,2,4,5</math>
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<math>3,1,2,5,4</math>
 +
<math>3,1,4,2,5</math>
 +
<math>3,1,4,5,2</math>
 +
<math>3,1,5,2,4</math>
 +
<math>3,1,5,4,2</math>
 +
<math>3,2,1,4,5</math>
 +
<math>3,2,1,5,4</math>
 +
<math>3,2,4,1,5</math>
 +
<math>3,2,4,5,1</math>
 +
<math>3,2,5,1,4</math>
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<math>3,2,5,4,1</math>
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<math>3,4,1,2,5</math>
 +
<math>3,4,1,5,2</math>
 +
<math>3,4,2,1,5</math>
 +
<math>3,4,2,5,1</math>
 +
<math>3,4,5,1,2</math>
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<math>3,4,5,2,1</math>
 +
<math>3,5,1,2,4</math>
 +
<math>3,5,1,4,2</math>
 +
<math>3,5,2,1,4</math>
 +
<math>3,5,2,4,1</math>
 +
<math>3,5,4,1,2</math>
 +
<math>3,5,4,2,1</math>
 +
<math>4,1,2,3,5</math>
 +
<math>4,1,2,5,3</math>
 +
<math>4,1,3,2,5</math>
 +
<math>4,1,3,5,2</math>
 +
<math>4,1,5,2,3</math>
 +
<math>4,1,5,3,2</math>
 +
<math>4,2,1,3,5</math>
 +
<math>4,2,1,5,3</math>
 +
<math>4,2,3,1,5</math>
 +
<math>4,2,3,5,1</math>
 +
<math>4,2,5,1,3</math>
 +
<math>4,2,5,3,1</math>
 +
<math>4,3,1,2,5</math>
 +
<math>4,3,1,5,2</math>
 +
<math>4,3,2,1,5</math>
 +
<math>4,3,2,5,1</math>
 +
<math>4,3,5,1,2</math>
 +
<math>4,3,5,2,1</math>
 +
<math>4,5,1,2,3</math>
 +
<math>4,5,1,3,2</math>
 +
<math>4,5,2,1,3</math>
 +
<math>4,5,2,3,1</math>
 +
<math>4,5,3,1,2</math>
 +
<math>4,5,3,2,1</math>
 +
<math>5,1,2,3,4</math>
 +
<math>5,1,2,4,3</math>
 +
<math>5,1,3,2,4</math>
 +
<math>5,1,3,4,2</math>
 +
<math>5,1,4,2,3</math>
 +
<math>5,1,4,3,2</math>
 +
<math>5,2,1,3,4</math>
 +
<math>5,2,1,4,3</math>
 +
<math>5,2,3,1,4</math>
 +
<math>5,2,3,4,1</math>
 +
<math>5,2,4,1,3</math>
 +
<math>5,2,4,3,1</math>
 +
<math>5,3,1,2,4</math>
 +
<math>5,3,1,4,2</math>
 +
<math>5,3,2,1,4</math>
 +
<math>5,3,2,4,1</math>
 +
<math>5,3,4,1,2</math>
 +
<math>5,3,4,2,1</math>
 +
<math>5,4,1,2,3</math>
 +
<math>5,4,1,3,2</math>
 +
<math>5,4,2,1,3</math>
 +
<math>5,4,2,3,1</math>
 +
<math>5,4,3,1,2</math>
 +
<math>5,4,3,2,1</math>
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[/hide]
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Some quick work gets us <math>\boxed{\text{D: }32}</math> permutations that work. ~contactbibliophile

Revision as of 15:08, 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. [hide=The Cases] $1,2,3,4,5$ $1,2,3,5,4$ $1,2,4,3,5$ $1,2,4,5,3$ $1,2,5,3,4$ $1,2,5,4,3$ $1,3,2,4,5$ $1,3,2,5,4$ $1,3,4,2,5$ $1,3,4,5,2$ $1,3,5,2,4$ $1,3,5,4,2$ $1,4,2,3,5$ $1,4,2,5,3$ $1,4,3,2,5$ $1,4,3,5,2$ $1,4,5,2,3$ $1,4,5,3,2$ $1,5,2,3,4$ $1,5,2,4,3$ $1,5,3,2,4$ $1,5,3,4,2$ $1,5,4,2,3$ $1,5,4,3,2$ $2,1,3,4,5$ $2,1,3,5,4$ $2,1,4,3,5$ $2,1,4,5,3$ $2,1,5,3,4$ $2,1,5,4,3$ $2,3,1,4,5$ $2,3,1,5,4$ $2,3,4,1,5$ $2,3,4,5,1$ $2,3,5,1,4$ $2,3,5,4,1$ $2,4,1,3,5$ $2,4,1,5,3$ $2,4,3,1,5$ $2,4,3,5,1$ $2,4,5,1,3$ $2,4,5,3,1$ $2,5,1,3,4$ $2,5,1,4,3$ $2,5,3,1,4$ $2,5,3,4,1$ $2,5,4,1,3$ $2,5,4,3,1$ $3,1,2,4,5$ $3,1,2,5,4$ $3,1,4,2,5$ $3,1,4,5,2$ $3,1,5,2,4$ $3,1,5,4,2$ $3,2,1,4,5$ $3,2,1,5,4$ $3,2,4,1,5$ $3,2,4,5,1$ $3,2,5,1,4$ $3,2,5,4,1$ $3,4,1,2,5$ $3,4,1,5,2$ $3,4,2,1,5$ $3,4,2,5,1$ $3,4,5,1,2$ $3,4,5,2,1$ $3,5,1,2,4$ $3,5,1,4,2$ $3,5,2,1,4$ $3,5,2,4,1$ $3,5,4,1,2$ $3,5,4,2,1$ $4,1,2,3,5$ $4,1,2,5,3$ $4,1,3,2,5$ $4,1,3,5,2$ $4,1,5,2,3$ $4,1,5,3,2$ $4,2,1,3,5$ $4,2,1,5,3$ $4,2,3,1,5$ $4,2,3,5,1$ $4,2,5,1,3$ $4,2,5,3,1$ $4,3,1,2,5$ $4,3,1,5,2$ $4,3,2,1,5$ $4,3,2,5,1$ $4,3,5,1,2$ $4,3,5,2,1$ $4,5,1,2,3$ $4,5,1,3,2$ $4,5,2,1,3$ $4,5,2,3,1$ $4,5,3,1,2$ $4,5,3,2,1$ $5,1,2,3,4$ $5,1,2,4,3$ $5,1,3,2,4$ $5,1,3,4,2$ $5,1,4,2,3$ $5,1,4,3,2$ $5,2,1,3,4$ $5,2,1,4,3$ $5,2,3,1,4$ $5,2,3,4,1$ $5,2,4,1,3$ $5,2,4,3,1$ $5,3,1,2,4$ $5,3,1,4,2$ $5,3,2,1,4$ $5,3,2,4,1$ $5,3,4,1,2$ $5,3,4,2,1$ $5,4,1,2,3$ $5,4,1,3,2$ $5,4,2,1,3$ $5,4,2,3,1$ $5,4,3,1,2$ $5,4,3,2,1$ [/hide] Some quick work gets us $\boxed{\text{D: }32}$ permutations that work. ~contactbibliophile

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