Try this beautiful Problem based on Enumeration appeared in AMC 10A 2021, 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?
Permutation
Enumeration
Combinatorics
An Excursion in Mathematics
AMC 10A 2021 Problem 20
32
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?
