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?

Other useful links

AMC-AIME Program at Cheenta

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?

Other useful links

AMC-AIME Program at Cheenta

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