Difference between revisions of "2021 AMC 10A Problems/Problem 20"
MRENTHUSIASM (talk | contribs) m (→Solution 3 (similar to solution 2)) |
(→Solution (bashing)) |
||
| Line 4: | Line 4: | ||
==Solution (bashing)== | ==Solution (bashing)== | ||
| − | We write out the <math> | + | We write out the <math>32</math> cases. |
These cases are the ones that work: | These cases are the ones that work: | ||
<math>13254,14253,14352,15243,15342,21435,21534,23154,24153,24351,25143,25341,\linebreak | <math>13254,14253,14352,15243,15342,21435,21534,23154,24153,24351,25143,25341,\linebreak | ||
Revision as of 13:54, 12 February 2021
Contents
Problem
In how many ways can the sequence 
be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
Solution (bashing)
We write out the 
cases.
These cases are the ones that work:
We count these out and get 
permutations that work. ~contactbibliophile
Solution 2 (Casework)
Reading the terms from left to right, we have two cases:
(1) 
(2) 
(
stands for increase and 
stands for decrease.)
For Case (1), note that for the 2nd and 4th terms, one of which must be a 5, and the other one must be a 3 or 4. We have four scenarios:
_3_5_
_5_3_
_4_5_
_5_4_
For the first scenario, the first two blanks must be 1 and 2 in some order, and the last blank must be a 4, for a total of 2 possibilities. Similarly, the second scenario also has 2 possibilities.
For the third scenario, there are no restrictions for the numbers 1, 2, and 3. So, we have 
possibilities. Similarly, the fourth scenario also has 6 possibilities.
Together, Case (1) has 
possibilities. By symmetry, Case (2) also has 
possibilities. Together, the answer is 
This problem is a little similar to the 2004 AIME I Problem 6: https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_6
~MRENTHUSIASM
Solution 3 (similar to solution 2)
Like Solution 2, we have two cases. Due to symmetry, we just need to count one of the cases. For the purpose of this solution, we will be doing . Instead of starting with 5, we start with 1.
There are two ways to place it:
_1_ _ _
_ _ _1_
Now we place 2, it can either be next to 1 and on the outside, or is place in where 1 would go in the other case. So now we have another two "sub case":
_1_2_(case 1)
21_ _ _(case 2)
There are 3! ways to arrange the rest for case 1, since there is no restriction.
For case 2, we need to consider how many ways to arrange 3,4,5 in a a>b<c fashion. It should seem pretty obvious that b has to be 3, so there will be 2! way to put 4 and 5.
Now we find our result, times 2 for symmetry, times 2 for placement of 1 and times (3!+2!) for the two different cases for placement of 2. This give us 
.
Video Solution by OmegaLearn (Using PIE - Principal of Inclusion Exclusion)
~ pi_is_3.14
See Also
| 2021 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
