2020 AIME I Problems/Problem 5

Problem

Six cards numbered through are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order.

Solution 1

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 through , there are five other spots to put them, so we get . However, we overcount some cases. Take the example of . We overcount this case because we can remove the or the . Therefore, any cases with two adjacent numbers swapped is overcounted, so we subtract cases (namely, ,) to get , but we have to add back one more for the original case, . Therefore, there are cases. Multiplying by gives the desired answer, .

-molocyxu

Video Solution:

https://www.youtube.com/watch?v=E6YJh7vsLPU

Solution 2 (Inspired by 2018 CMIMC combo round)

Similar to above, a correspondence between ascending and descending is established by subtracting each number from .

We note that the given condition is equivalent to "cycling" for a contiguous subset of it. For example,

It's not hard to see that no overcount is possible, and that the cycle is either "right" or "left." Therefore, we consider how many elements we flip by. If we flip or such elements, then there is one way to cycle them. Otherwise, we have ways. Therefore, the total number of ascending is , and multiplying by two gives ~awang11

Solution 3

Similarly to above, we find the number of ascending arrangements and multiply by 2.

We can choose cards to be the ascending cards, therefore leaving places to place the remaining card. There are to do this. However, since the problem is asking for the number of arrangements, we overcount cases such as . Notice that the only arrangements that overcount are (case 1) or if two adjacent numbers of are switched (case 2).

This arrangement is counted times. Each time it is counted for any of the numbers selected. Therefore we need to subtract cases of overcounting.

Each time adjacent numbers of switched, there is one overcount. For example, if we have , both or could be removed. Since there are possible switches, we need to subtract cases of overcounting.

Therefore, we have total arrangements of ascending numbers. We multiply by two (for descending) to get the answer of -PCChess

Solution 4 (No overcounting)

Like in previous solutions, we will count the number of ascending arrangements and multiply by 2.

First, consider the arrangement 1-2-3-4-5-6. That gives us 1 arrangement which works.

Next, we can switch two adjacent cards. There are 5 ways to pick two adjacent cards, so this gives us 5 arrangements.

Now, we can "cycle" 3 adjacent cards. For example, 1-2-3 becomes 2-3-1 which becomes 3-1-2. There are 4 ways to pick a set of 3 adjacent cards, so this gives us 4x2=8 arrangements.

Cycling 4 adjacent cards, we get the new arrangements 2-3-4-1 (which works,) 3-4-1-2 (which doesn't work,) and 4-1-2-3 (which does work.) We get 6 arrangements.

Similarly, when cycling 5 cards, we find 2x2=4 arrangements, and when cycling 6 cards, we find 2x1=2 arrangements.

Adding, we figure out that there are 1+5+8+6+4+2=26 ascending arrangements. Multiplying by 2, we get the answer -i8Pie

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.