Difference between revisions of "2020 AIME I Problems/Problem 5"
MRENTHUSIASM (talk | contribs) m (→Solution 9 (Illustration)) |
Arcticturn (talk | contribs) (→Solution 10) |
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WLOG let us consider ascending order | WLOG let us consider ascending order | ||
− | Case 1: after we take out 1, the rest will be in ascending order | + | Case 1: after we take out 1, the rest will be in ascending order: |
+ | <center>2 3 4 5 6</center> | ||
+ | Notice that 1 can be tucked in any one of the 6 spaces, thus there are 6 scenarios. | ||
− | + | Case 2: after we take out 2, the rest will be in ascending order: | |
− | + | <center>1 3 4 5 6</center> | |
− | + | Notice that if we put 2 next to 1 (to the right or to the left of 1), it will be an overcount, so there are only 4 cases for 2. | |
− | |||
− | Case 2: after we take out 2, the rest will be in ascending order | ||
− | |||
− | 1 3 4 5 6 | ||
− | |||
− | |||
It is easy to see that this is the same for 3, 4, 5, and 6. | It is easy to see that this is the same for 3, 4, 5, and 6. | ||
− | Thus, in total, we have | + | Thus, in total, we have <cmath>(6+4\times5)\times2=\boxed{052}</cmath> |
− | <cmath>(6+4\times5)\times2=\boxed{052}</cmath> | ||
~Adali | ~Adali | ||
Line 90: | Line 85: | ||
Based on the card to be inserted, we have six cases. As shown below, the red squares indicate the possible positions to insert the sixth card. | Based on the card to be inserted, we have six cases. As shown below, the red squares indicate the possible positions to insert the sixth card. | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | unitsize(7mm); | ||
− | + | void drawSquare(real x, real y) | |
+ | { | ||
+ | draw((x+0.5,y+0.5)--(x-0.5,y+0.5)--(x-0.5,y-0.5)--(x+0.5,y-0.5)--cycle,red); | ||
+ | } | ||
+ | label("$2$",(0.5,8)); | ||
+ | label("$3$",(2.5,8)); | ||
+ | label("$4$",(4.5,8)); | ||
+ | label("$5$",(6.5,8)); | ||
+ | label("$6$",(8.5,8)); | ||
+ | label("Insert $1.$",(13.5,8),red); | ||
+ | drawSquare(-0.5,8); | ||
+ | drawSquare(1.5,8); | ||
+ | drawSquare(3.5,8); | ||
+ | drawSquare(5.5,8); | ||
+ | drawSquare(7.5,8); | ||
+ | drawSquare(9.5,8); | ||
+ | |||
+ | label("$1$",(0.5,6.5)); | ||
+ | label("$3$",(2.5,6.5)); | ||
+ | label("$4$",(4.5,6.5)); | ||
+ | label("$5$",(6.5,6.5)); | ||
+ | label("$6$",(8.5,6.5)); | ||
+ | label("Insert $2.$",(13.5,6.5),red); | ||
+ | drawSquare(3.5,6.5); | ||
+ | drawSquare(5.5,6.5); | ||
+ | drawSquare(7.5,6.5); | ||
+ | drawSquare(9.5,6.5); | ||
+ | |||
+ | label("$1$",(0.5,5)); | ||
+ | label("$2$",(2.5,5)); | ||
+ | label("$4$",(4.5,5)); | ||
+ | label("$5$",(6.5,5)); | ||
+ | label("$6$",(8.5,5)); | ||
+ | label("Insert $3.$",(13.5,5),red); | ||
+ | drawSquare(-0.5,5); | ||
+ | drawSquare(5.5,5); | ||
+ | drawSquare(7.5,5); | ||
+ | drawSquare(9.5,5); | ||
+ | |||
+ | label("$1$",(0.5,3.5)); | ||
+ | label("$2$",(2.5,3.5)); | ||
+ | label("$3$",(4.5,3.5)); | ||
+ | label("$5$",(6.5,3.5)); | ||
+ | label("$6$",(8.5,3.5)); | ||
+ | label("Insert $4.$",(13.5,3.5),red); | ||
+ | drawSquare(-0.5,3.5); | ||
+ | drawSquare(1.5,3.5); | ||
+ | drawSquare(7.5,3.5); | ||
+ | drawSquare(9.5,3.5); | ||
+ | |||
+ | label("$1$",(0.5,2)); | ||
+ | label("$2$",(2.5,2)); | ||
+ | label("$3$",(4.5,2)); | ||
+ | label("$4$",(6.5,2)); | ||
+ | label("$6$",(8.5,2)); | ||
+ | label("Insert $5.$",(13.5,2),red); | ||
+ | drawSquare(-0.5,2); | ||
+ | drawSquare(1.5,2); | ||
+ | drawSquare(3.5,2); | ||
+ | drawSquare(9.5,2); | ||
+ | |||
+ | label("$1$",(0.5,0.5)); | ||
+ | label("$2$",(2.5,0.5)); | ||
+ | label("$3$",(4.5,0.5)); | ||
+ | label("$4$",(6.5,0.5)); | ||
+ | label("$5$",(8.5,0.5)); | ||
+ | label("Insert $6.$",(13.5,0.5),red); | ||
+ | drawSquare(-0.5,0.5); | ||
+ | drawSquare(1.5,0.5); | ||
+ | drawSquare(3.5,0.5); | ||
+ | drawSquare(5.5,0.5); | ||
+ | </asy> | ||
Note that all arrangements of the six cards are distinct. | Note that all arrangements of the six cards are distinct. | ||
Line 98: | Line 167: | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 10 == | ||
+ | We first realize that as long as we have an ascending sequence of 5 numbers, we can just plug in a 6th to make a sequence that works. For example, if we have 12345, we can plug in a 6 in any of 6 spaces, before 1, between 1 and 2, and so on to after 5. We can also realize that this is completely symmetrical if the sequence is in descending order. For example, we could have 54321, and we could plug in a 6 in 6 of the spaces. For the total number of combinations, we have 6 ascending cases multiplied by the 6 places that can hold whatever number is missing, and multiply by 2 because there is descending and ascending number cases. | ||
+ | |||
+ | But wait, what if we have 12345 and plug in a 6 at the end and 12346 and plug in a 5 at the 5th spot? We have overcounted, so we need to subtract off the identical pairs. Assume that one of them is the "right" combination. That means that there are 1 "right" combination because all the rest will have 2 of their combinations taken by the previous one. For example, if I have 12356 and 12346, I would count those 123456 as 12356's instead of 12346's. Therefore, we have 6+4+4+4+4+4 = 26 combinations of ascending order, and since we need to count those of descending order as well, we have <math>26\cdot2=\boxed{052}.</math> | ||
+ | |||
+ | Side note: We could have just realized that the first part was unnecessary, but it does help you understand the problem better. | ||
+ | |||
+ | ~Arcticturn | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 10:43, 26 September 2021
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Inspired by 2018 CMIMC Combo Round)
- 4 Solution 3
- 5 Solution 4 (No Overcounting)
- 6 Solution 5 (Official MAA 1)
- 7 Solution 6 (Official MAA 2)
- 8 Solution 7 (Casework)
- 9 Solution 8 (Symmetry and Case Study)
- 10 Solution 9 (Illustration)
- 11 Solution 10
- 12 Video Solution
- 13 Video Solution
- 14 Video Solution
- 15 See Also
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
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
Solution 5 (Official MAA 1)
First count the number of permutations of the cards such that if one card is removed, the remaining cards will be in ascending order. There is such permutation where all the cards appear in order: There are such permutations where two adjacent cards are interchanged, as in The other such permutations arise from removing one card from and placing it in a position at least two away from its starting location. There are such positions to place each of the cards numbered and and such positions for each of the cards numbered and This accounts for permutations. Thus there are permutations where one card can be removed so that the remaining cards are in ascending order. There is an equal number of permutations that result in the cards' being in descending order. This gives the total .
Solution 6 (Official MAA 2)
More generally, suppose there are cards numbered arranged in ascending order. If any one of the cards is removed and placed in one of the positions in the arrangement, the resulting permutation will have the property that one card can be removed so that the remaining cards are in ascending order. This accounts for permutations. However, the original ascending order has been counted times, and each order that arises by switching two neighboring cards has been counted twice. Hence the number of arrangements where one card can be removed resulting in the remaining cards' being in ascending order is When , this is , and the final answer is .
Solution 7 (Casework)
For ascending, if the goes in anything but the first two slots, the rest of the numbers have to go in ascending from , which are cases if there are cards. If goes in the second spot, then you can put any of the rest in the first slot but then the rest are determined, so in the case of cards, that gives more. If goes in the first slot, that means that you are doing the same problem with cards. So the recursion is . There's and , so you get , , , and . Or you can see that . We double to account for descending and get .
~ahclark11
Solution 8 (Symmetry and Case Study)
First, we know that ascending order and descending order are symmetrical to each other (namely, if we get 132456 where after we take out 3, it will be one scenario; and if we flip it and write 654231, it will be another scenario)
Thus, we only need to consider either descending or ascending and then times 2. WLOG let us consider ascending order
Case 1: after we take out 1, the rest will be in ascending order:
Notice that 1 can be tucked in any one of the 6 spaces, thus there are 6 scenarios.
Case 2: after we take out 2, the rest will be in ascending order:
Notice that if we put 2 next to 1 (to the right or to the left of 1), it will be an overcount, so there are only 4 cases for 2.
It is easy to see that this is the same for 3, 4, 5, and 6.
Thus, in total, we have ~Adali
Solution 9 (Illustration)
We start with five cards in ascending order, then insert the sixth card to obtain a valid arrangement.
Based on the card to be inserted, we have six cases. As shown below, the red squares indicate the possible positions to insert the sixth card. Note that all arrangements of the six cards are distinct.
There are arrangements in which removing one card will leave the remaining five cards in ascending order. By symmetry, there are arrangements in which removing one card will leave the remaining five cards in descending order. So, the answer is
~MRENTHUSIASM
Solution 10
We first realize that as long as we have an ascending sequence of 5 numbers, we can just plug in a 6th to make a sequence that works. For example, if we have 12345, we can plug in a 6 in any of 6 spaces, before 1, between 1 and 2, and so on to after 5. We can also realize that this is completely symmetrical if the sequence is in descending order. For example, we could have 54321, and we could plug in a 6 in 6 of the spaces. For the total number of combinations, we have 6 ascending cases multiplied by the 6 places that can hold whatever number is missing, and multiply by 2 because there is descending and ascending number cases.
But wait, what if we have 12345 and plug in a 6 at the end and 12346 and plug in a 5 at the 5th spot? We have overcounted, so we need to subtract off the identical pairs. Assume that one of them is the "right" combination. That means that there are 1 "right" combination because all the rest will have 2 of their combinations taken by the previous one. For example, if I have 12356 and 12346, I would count those 123456 as 12356's instead of 12346's. Therefore, we have 6+4+4+4+4+4 = 26 combinations of ascending order, and since we need to count those of descending order as well, we have
Side note: We could have just realized that the first part was unnecessary, but it does help you understand the problem better.
~Arcticturn
Video Solution
https://www.youtube.com/watch?v=5iwdFd2OLKM&list=PLLCzevlMcsWN9y8YI4KNPZlhdsjPTlRrb&index=5 ~ MathEx
Video Solution
https://www.youtube.com/watch?v=E6YJh7vsLPU
Video Solution
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.