Difference between revisions of "2011 AMC 12A Problems/Problem 16"
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6.) Now, we can multiply these three paths and add them: | 6.) Now, we can multiply these three paths and add them: | ||
− | <math>(6\times1\times5\times5\times4)+(6\times5\times4\times4\times4)+(6\times5\times1\times4\times5)=600+1920+600=3120</math> | + | <math>(6\times1\times5\times5\times4)+(6\times5\times4\times4\times4)+(6\times5\times1\times4\times5) |
+ | =600+1920+600=3120</math> | ||
7.) Our answer is <math>C</math>! | 7.) Our answer is <math>C</math>! | ||
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We get <math>6\times5\times4\times3=360</math>. | We get <math>6\times5\times4\times3=360</math>. | ||
− | However, there are 5 different locations the pair could be at. Therefore we get <math>360\times5=1800</math> possibilities for | + | However, there are 5 different locations the pair could be at. Therefore we get <math>360\times5=1800</math> possibilities for one pair. |
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We get <math>6\times5\times4=120</math>. | We get <math>6\times5\times4=120</math>. | ||
− | Once again, there are 5 different rotations in the pentagon that we must account for. Therefore we get <math>120\times5=600</math>. | + | Once again, there are 5 different rotations in the pentagon that we must account for. Therefore we get <math>120\times5=600</math> possibilities for two pairs. |
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Solution by gsaelite | Solution by gsaelite | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | This solution essentially explains other ways of thinking about the cases stated in Solution 2. | ||
+ | |||
+ | Case 1: | ||
+ | |||
+ | <math>{6\choose5} \cdot 5!</math> | ||
+ | |||
+ | 5 colors need to be chosen from the group of 6. Those 5 colors have 5! distinct arrangements on the pentagon's vertices. | ||
+ | |||
+ | Case 2: | ||
+ | |||
+ | <math>{6\choose4} \cdot4\cdot5\cdot3!</math> | ||
+ | |||
+ | 4 colors need to be chosen from the group of 6. Out of those 4 colors, one needs to be chosen to form a pair of 2 identical colors. Then, for arranging this layout onto the pentagon, one of the five sides (same thing as pair of adjacent vertices) of the pentagon needs to be established for the pair. The remaining 3 unique colors can be arranged 3! different ways on the remaining 3 vertices. | ||
+ | |||
+ | Case 3: | ||
+ | |||
+ | <math>{6\choose3} \cdot {3\choose2} \cdot5\cdot2</math> | ||
+ | |||
+ | 3 colors need to be chosen from the group of 6. Out of those 3 colors, 2 need to be chosen to be the pairs. Then, for arranging this layout onto the pentagon, start out by thinking about the 1 color that is not part of a pair, as it makes things easier. It can be any one of the 5 vertices of the pentagon. The remaining 2 pairs of colors can only be arranged 2 ways on the remaining 4 vertices. | ||
+ | |||
+ | Solving each case and adding them up gets you 3120. <math>\boxed{C}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/FThly7dRBIE | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
== See also == | == See also == |
Latest revision as of 13:41, 7 September 2021
Problem
Each vertex of convex pentagon is to be assigned a color. There are colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
Solution
We can do some casework when working our way around the pentagon from to . At each stage, there will be a makeshift diagram.
1.) For , we can choose any of the 6 colors.
A : 6
2.) For , we can either have the same color as , or any of the other 5 colors. We do this because each vertex of the pentagon is affected by the 2 opposite vertices, and will be affected by both and .
A : 6 B:1 B:5
3.) For , we cannot have the same color as . Also, we can have the same color as ( will be affected), or any of the other 4 colors. Because can't be the same as , it can't be the same as if is the same as , so it can be any of the 5 other colors.
A : 6 B:1 B:5 C:5 C:4 C:1
4.) is affected by and . If they are the same, then can be any of the other 5 colors. If they are different, then can be any of the (6-2)=4 colors.
A : 6 B:1 B:5 C:5 C:4 C:1 D:5 D:4 D:4
5.) is affected by and . If they are the same, then can be any of the other 5 colors. If they are different, then can be any of the (6-2)=4 colors.
A : 6 B:1 B:5 C:5 C:4 C:1 D:5 D:4 D:4 E:4 E:4 E:5
6.) Now, we can multiply these three paths and add them:
7.) Our answer is !
Solution 2
Right off the bat, we can analyze three things:
1.) There can only be two of the same color on the pentagon.
2.) Any pair of the same color can only be next to each other on the pentagon.
3.) There can only be two different pairs of same colors on the pentagon at once.
Now that we know this, we can solve the problem by using three cases: no same color pairs, one same color pair, and two same color pairs.
1.) If there are no color pairs, it is a simple permutation: six different colors in five different spots. We count cases. No rotation is necessary because all permutations are accounted for.
2.)If there is one color pair, we must count 6 possibilities for the pair(as one element), 5 for the third vertex, 4 for the fourth vertex, and 3 for the fifth vertex.
We get .
However, there are 5 different locations the pair could be at. Therefore we get possibilities for one pair.
3.)If there are two color pairs, we must count 6 possibilities for the first pair(as one element), 5 possibilities for the next pair(as one element), and 4 possibilities for the final vertex.
We get .
Once again, there are 5 different rotations in the pentagon that we must account for. Therefore we get possibilities for two pairs.
5.) If we add all of three cases together, we get . The answer is .
Solution by gsaelite
Solution 3
This solution essentially explains other ways of thinking about the cases stated in Solution 2.
Case 1:
5 colors need to be chosen from the group of 6. Those 5 colors have 5! distinct arrangements on the pentagon's vertices.
Case 2:
4 colors need to be chosen from the group of 6. Out of those 4 colors, one needs to be chosen to form a pair of 2 identical colors. Then, for arranging this layout onto the pentagon, one of the five sides (same thing as pair of adjacent vertices) of the pentagon needs to be established for the pair. The remaining 3 unique colors can be arranged 3! different ways on the remaining 3 vertices.
Case 3:
3 colors need to be chosen from the group of 6. Out of those 3 colors, 2 need to be chosen to be the pairs. Then, for arranging this layout onto the pentagon, start out by thinking about the 1 color that is not part of a pair, as it makes things easier. It can be any one of the 5 vertices of the pentagon. The remaining 2 pairs of colors can only be arranged 2 ways on the remaining 4 vertices.
Solving each case and adding them up gets you 3120.
Video Solution
~IceMatrix
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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