2021 AMC 10A Problems/Problem 20
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 1 (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:
(
stands for increase and 
stands for decrease.)
For 
note that for the second and fourth terms, one of which must be a 
and the other one must be a 
or 
We have four sub-cases:
For 
the first two blanks must be 
and 
in some order, and the last blank must be a 
for a total of possibilities. Similarly, 
also has possibilities.
For 
there are no restrictions for the numbers 
and 
So, we have 
possibilities. Similarly, 
also has 
possibilities.
Together, 
has 
possibilities. By symmetry, 
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 
.
~~Xhte
Solution 4: Symmetry
We only need to find # of rearrangements when 5 is the 4th digit and 5th digit. Find the total, and multiply by 2. Then we can get the answer by adding the case when 5 is the third digit.
Case 1: 5 is the 5th digit. __ __ __ __ 5""" Then 4 can only be either 1st digit or the 3rd digit.
4 __ __ __ 5, then the only way is that 3 is the 3rd digit, so it can be either 231 or 132, give us 2 results.
__ __ 4 __ 5, then the 1st digit must be 2 or 3, 2 gives us 1 way, and 3 gives us 2 ways. (Can't be 1 because the first digit would increasing). Therefore, 4 in the middle and 5 in the last would result in 3 ways.
"""Case 2: 5 is the fourth digit. __ __ __ 5 __"""
Then the last digit can be all of the 4 numbers 1, 2, 3, and 4. Let's say if the last digit is 4, then the 2nd digit would be the largest for the remaining digits to prevent increasing order or decreasing order. Then the remaining two are interchangable, give us 2! ways. All of the 4 can work, so case 2 would result in 2!+2!+2!+2!=8 ways.
"""Case 3: 5 is in the middle __ __ 5 __ __"""
Then there are only two cases: 1. 42513, then 4 and 3 are interchangable, which results in 2!*2!. Or it can be 43512, then 4 and 2 are interchangable, but it can not be 23514, so there can only be 2 possible ways: 43512, 21534.
Therefore, case 3 would result in 4+2=6 ways.
8+3+2=13, so the total ways for case 1 and case 2 with both increasing and decreasing would be 13*2 = 26.
26+6=32.
~~Michael595
Video Solution by OmegaLearn (Using PIE - Principle of Inclusion Exclusion)
~ pi_is_3.14
Video Solution by Power of Logic (Michael595) (Using idea of symmetrically counting)
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 | ||
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