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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

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

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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