2021 AMC 10A Problems/Problem 20
Contents
- 1 Problem
- 2 Solution 1 (Enumeration)
- 3 Solution 2 (Enumeration by Symmetry)
- 4 Solution 3 (Casework on the Consecutive Digits)
- 5 Solution 4 (Casework Similar to Solution 3)
- 6 Solution 5 (Casework on the Position of 5)
- 7 Solution 6 (Overcounting)
- 8 Video Solution by OmegaLearn (Using PIE - Principle of Inclusion Exclusion)
- 9 Video Solution by Power of Logic (Using Idea of Symmetrically Counting)
- 10 Video Solution by TheBeautyofMath
- 11 See Also
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 (Enumeration)
We write out the cases, then filter out the valid ones:
We count these out and get permutations that work.
~contactbibliophile
Solution 2 (Enumeration by Symmetry)
By symmetry with respect to note that is a valid sequence if and only if is a valid sequence. We enumerate the valid sequences that start with or as shown below:
There are valid sequences that start with or By symmetry, there are valid sequences that start with or So, the answer is
~MRENTHUSIASM (inspired by Snowfan)
Solution 3 (Casework on the Consecutive Digits)
Reading the terms from left to right, we have two cases for the consecutive digits, where means increase and means decrease:
For note that for the second and fourth terms, one term must be and the other term must be either or We have four subcases:
For the first two blanks must be and in some order, and the last blank must be So, we get possibilities. Similarly, also has possibilities.
For there are no restrictions for the numbers and So, we get possibilities. Similarly, also has possibilities.
Together, has possibilities. By symmetry, also has possibilities.
Finally, the answer is
Remark
This problem is somewhat similar to 2004 AIME I Problem 6.
~MRENTHUSIASM
Solution 4 (Casework Similar to Solution 3)
Like Solution 3, 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 5 (Casework on the Position of 5)
We only need to find the # 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 : 5 is the 5th digit. __ __ __ __ 5
Then can only be either 1st digit or the 3rd digit.
4 __ __ __ 5, then the only way is that is the 3rd digit, so it can be either or , give us results.
__ __ 4 __ 5, then the 1st digit must be or , gives us way, and gives us ways. (Can't be because the first digit would increasing). Therefore, in the middle and in the last would result in ways.
Case : is the fourth digit. __ __ __ 5 __
Then the last digit can be all of the 4 numbers , , , and . Let's say if the last digit is , then the 2nd digit would be the largest for the remaining digits to prevent increasing order or decreasing order. Then the remaining two are interchangeable, give us ways. All of the can work, so case would result in ways.
Case : is in the middle. __ __ 5 __ __
Then there are only two cases: 1. , then 4 and 3 are interchangeable, which results in . Or it can be , then 4 and 2 are interchangeable, but it can not be , so there can only be 2 possible ways: , .
Therefore, case 3 would result in ways.
, so the total ways for case 1 and case 2 with both increasing and decreasing would be
Finally, we have
~Michael595
Solution 6 (Overcounting)
First, we list the triples that are invalid:
543, 542, 541, 532, 531, 521, 432, 431, 321
By symmetry, there are the same amount of increasing triplets as there are decreasing ones. This yields 18 invalid 3 digit permutations in total.
Suppose the triplet is ABC and the other 2 digits are X and Y. We then have 3 ways to arrange a triplet with 2 other digits.
ABCXY, XABCY, XYABC
X and Y can be arranged 2 ways.
XY, YX
This produces 18*3*2=108 permutations of invalid results. We have 5! ways to arrange 5 numbers so 120-108=12.
Now, we must account for overcounting. For example, when 543 is counted, it only registers as one invalid permutation but in fact, it is 3 whole invalid permutations. We then complete this for the rest of the list:
54321 has 543, 432, and 321
54213 has 542 and 421
54123 has 541 and 123
53214 has 532 and 321
53124 has 531 and 124
52134 has 521 and 134
43215 has 432 and 321
43125 has 431 and 125
32145 has 321 and 145
This produces 19 values that we have overcounted but this value itself is also overcounted. We already counted 9 of the terms. This brings the final value of overcounted terms down to 10 for the decreasing triplets. By symmetry, 10 increasing triplets were overcounted.
This gives us
~ Lukiebear
Video Solution by OmegaLearn (Using PIE - Principle of Inclusion Exclusion)
~ pi_is_3.14
Video Solution by Power of Logic (Using Idea of Symmetrically Counting)
Video Solution by TheBeautyofMath
~IceMatrix
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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