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AMC 10A 2021 I Problem 20 | Enumeration

Try this beautiful Problem based on Enumeration appeared in AMC 10A 2021, Problem 20.

AMC 10A 2021 I Problem 20


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?

  • 10
  • 18
  • 24
  • 32
  • 44

Key Concepts


Permutation

Enumeration

Combinatorics

Suggested Book | Source | Answer


An Excursion in Mathematics

AMC 10A 2021 Problem 20

32

Try with Hints


We have 5 numbers with us.

So, how many permutations we can have with those numbers?

So, 5!=120 numbers can be made out of those 5 numbers.

Now we have to remember that we are restricted with the following condition -

no three consecutive terms are increasing and no three consecutive terms are decreasing.

Now make a list of the numbers which are satisfying the condition given among all 120 numbers we can have.

Now the list should be -

13254, 14253, 14352, 15243, 15342, 21435, 21534, 23154, 24153, 24351, 25143, 25341
31425, 31524, 32415, 32514, 34152, 34251, 35142, 35241, 41325, 41523, 42315, 42513,
43512, 45132, 45231, 51324, 51423, 52314, 52413, 53412.

Count how many permutations are there?

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Try this beautiful Problem based on Enumeration appeared in AMC 10A 2021, Problem 20.

AMC 10A 2021 I Problem 20


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?

  • 10
  • 18
  • 24
  • 32
  • 44

Key Concepts


Permutation

Enumeration

Combinatorics

Suggested Book | Source | Answer


An Excursion in Mathematics

AMC 10A 2021 Problem 20

32

Try with Hints


We have 5 numbers with us.

So, how many permutations we can have with those numbers?

So, 5!=120 numbers can be made out of those 5 numbers.

Now we have to remember that we are restricted with the following condition -

no three consecutive terms are increasing and no three consecutive terms are decreasing.

Now make a list of the numbers which are satisfying the condition given among all 120 numbers we can have.

Now the list should be -

13254, 14253, 14352, 15243, 15342, 21435, 21534, 23154, 24153, 24351, 25143, 25341
31425, 31524, 32415, 32514, 34152, 34251, 35142, 35241, 41325, 41523, 42315, 42513,
43512, 45132, 45231, 51324, 51423, 52314, 52413, 53412.

Count how many permutations are there?

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