Difference between revisions of "1997 AIME Problems/Problem 10"
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How many different complementary three-card sets are there? | How many different complementary three-card sets are there? | ||
− | == Solution == | + | __TOC__ |
− | + | == Solution 1 == | |
+ | *'''Case 1''': All three attributes are the same. This is impossible since sets contain distinct cards. | ||
+ | *'''Case 2''': Two of the three attributes are the same. There are <math>{3\choose 2}</math> ways to pick the two attributes in question. Then there are <math>3</math> ways to pick the value of the first attribute, <math>3</math> ways to pick the value of the second attribute, and <math>1</math> way to arrange the positions of the third attribute, giving us <math>{3\choose 2} \cdot 3 \cdot 3 = 27</math> ways. | ||
+ | *'''Case 3''': One of the three attributes are the same. There are <math>{3\choose 1}</math> ways to pick the one attribute in question, and then <math>3</math> ways to pick the value of that attribute. Then there are <math>3!</math> ways to arrange the positions of the next two attributes, giving us <math>{3\choose 1} \cdot 3 \cdot 3! = 54</math> ways. | ||
+ | *'''Case 4''': None of the three attributes are the same. We fix the order of the first attribute, and then there are <math>3!</math> ways to pick the ordering of the second attribute and <math>3!</math> ways to pick the ordering of the third attribute. This gives us <math>(3!)^2 = 36</math> ways. | ||
− | < | + | Adding the cases up, we get <math>27 + 54 + 36 = \boxed{117}</math>. |
− | We | + | == Solution 2 == |
+ | Let's say we have picked two cards. We now compare their attributes to decide how we can pick the third card to make a complement set. For each of the three attributes, should the two values be the same we have one option - choose a card with the same value for that attribute. Furthermore, should the two be different there is only one option- choose the only value that is remaining. In this way, every two card pick corresponds to exactly one set, for a total of <math>\binom{27}{2} = 27*13 = 351</math> possibilities. Note, however, that each set is generated by <math>{3\choose 2} = 3</math> pairs, so we've overcounted by a multiple of 3 and the answer is <math>\frac{351}{3} = \boxed{117}</math>. | ||
− | < | + | == Solution 3== |
+ | Treat the sets as ordered. Then for each of the three criterion, there are <math>3!=6</math> choices if the attribute is different and there are <math>3</math> choices is the attribute is the same. Thus all three attributes combine to a total of <math>(6+3)^3=729</math> possibilities. However if all three attributes are the same then the set must be composed of three cards that are the same, which is impossible. This takes out <math>3^3=27</math> possibilities. Notice that we have counted every set <math>3!=6</math> times by treating the set as ordered. the final solution is then <math>\frac{729-27}{6}=\boxed{117}</math> | ||
− | + | ==See Also== | |
+ | {{AIME box|year=1997|num-b=9|num-a=11}} | ||
− | + | [[Category:Intermediate Combinatorics Problems]] | |
− | + | {{MAA Notice}} | |
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Latest revision as of 02:21, 7 November 2021
Problem
Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of three cards from the deck is called complementary if all of the following statements are true:
i. Either each of the three cards has a different shape or all three of the card have the same shape.
ii. Either each of the three cards has a different color or all three of the cards have the same color.
iii. Either each of the three cards has a different shade or all three of the cards have the same shade.
How many different complementary three-card sets are there?
Solution 1
- Case 1: All three attributes are the same. This is impossible since sets contain distinct cards.
- Case 2: Two of the three attributes are the same. There are ways to pick the two attributes in question. Then there are ways to pick the value of the first attribute, ways to pick the value of the second attribute, and way to arrange the positions of the third attribute, giving us ways.
- Case 3: One of the three attributes are the same. There are ways to pick the one attribute in question, and then ways to pick the value of that attribute. Then there are ways to arrange the positions of the next two attributes, giving us ways.
- Case 4: None of the three attributes are the same. We fix the order of the first attribute, and then there are ways to pick the ordering of the second attribute and ways to pick the ordering of the third attribute. This gives us ways.
Adding the cases up, we get .
Solution 2
Let's say we have picked two cards. We now compare their attributes to decide how we can pick the third card to make a complement set. For each of the three attributes, should the two values be the same we have one option - choose a card with the same value for that attribute. Furthermore, should the two be different there is only one option- choose the only value that is remaining. In this way, every two card pick corresponds to exactly one set, for a total of possibilities. Note, however, that each set is generated by pairs, so we've overcounted by a multiple of 3 and the answer is .
Solution 3
Treat the sets as ordered. Then for each of the three criterion, there are choices if the attribute is different and there are choices is the attribute is the same. Thus all three attributes combine to a total of possibilities. However if all three attributes are the same then the set must be composed of three cards that are the same, which is impossible. This takes out possibilities. Notice that we have counted every set times by treating the set as ordered. the final solution is then
See Also
1997 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.