# Difference between revisions of "2020 AIME I Problems/Problem 5"

Note: Please do not post problems here until after the AIME.

## Solution

Realize that any sequence that works (ascending) can be reversed for descending, so we can just take the amount of sequences that satisfy the ascending condition and multiply by two.

If we choose any of the numbers $1$ through $6$, there are five other spots to put them, so we get $6 \cdot 5 = 30$. However, we overcount some cases. Take the example of $132456$. We overcount this case because we can remove the $3$ or the $2$. Therefore, any cases with two adjacent numbers swapped is overcounted, so we subtract $5$ cases (namely, $213456, 132456, 124356, 123546, 123465$,) to get $30-5=25$, but we have to add back one more for the original case, $123456$. Therefore, there are $26$ cases. Multiplying by $2$ gives the desired answer, $\boxed{052}$.

-molocyxu