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

Line 6: Line 6:
 
We write out the <math>120</math> cases.  
 
We write out the <math>120</math> cases.  
 
These cases are the ones that work:
 
These cases are the ones that work:
<math>\begin{align}
+
<math>13254,
1,3,2,5,4
+
14253,
1,4,2,5,3
+
14352,
1,4,3,5,2
+
15243,
1,5,2,4,3
+
15342,
1,5,3,4,2
+
21435,
2,1,4,3,5
+
21534,
2,1,5,3,4
+
23154,
2,3,1,5,4
+
24153,
2,4,1,5,3
+
24351,
2,4,3,5,1
+
25143,
2,5,1,4,3
+
25341,
2,5,3,4,1
+
31425,
3,1,4,2,5
+
31524,
3,1,5,2,4
+
32415,
3,2,4,1,5
+
32451,
3,2,5,1,4
+
34152,
3,4,1,5,2
+
34251,
3,4,2,5,1
+
35142,
3,5,1,4,2
+
35241,
3,5,2,4,1
+
41325,
4,1,3,2,5
+
41523,
4,1,5,2,3
+
42315,
4,2,3,1,5
+
42513,
4,2,5,1,3
+
43512,
4,3,5,1,2
+
45132,
4,5,1,3,2
+
45231,
4,5,2,3,1
+
51324,
5,1,3,2,4
+
51423,
5,1,4,2,3
+
52314,
5,2,3,1,4
+
52413,
5,2,4,1,3
+
53412,</math>
5,3,4,1,2
 
\end{align}</math>
 
 
We count these out and get <math>\boxed{\text{D: }32}</math> permutations that work. ~contactbibliophile
 
We count these out and get <math>\boxed{\text{D: }32}</math> permutations that work. ~contactbibliophile

Revision as of 15:30, 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: $13254, 14253, 14352, 15243, 15342, 21435, 21534, 23154, 24153, 24351, 25143, 25341, 31425, 31524, 32415, 32451, 34152, 34251, 35142, 35241, 41325, 41523, 42315, 42513, 43512, 45132, 45231, 51324, 51423, 52314, 52413, 53412,$ We count these out and get $\boxed{\text{D: }32}$ permutations that work. ~contactbibliophile