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Difference between revisions of "1997 AIME Problems/Problem 10"

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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:
 
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:
  
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i. Either each of the three cards has a different shape or all three of the card have the same shape.
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i. Either each of the three cards has a different shape or all three of the cards 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.
 
ii. Either each of the three cards has a different color or all three of the cards have the same color.
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How many different complementary three-card sets are there?
 
How many different complementary three-card sets are there?
  
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== Solution ==
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__TOC__
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We call these three types of complementary sets <math>A,B,C</math> respectively. What we are trying to find is
 
  
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<cmath>n(A\cup B\cup C)</cmath>
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== Solution 1 ==
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*'''Case 1''': All three attributes are the same. This is impossible since sets contain distinct cards.
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*'''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.
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*'''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.
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*'''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.
  
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We know this is equivalent to
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Adding the cases up, we get <math>27 + 54 + 36 = \boxed{117}</math>.
  
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<cmath>n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B \cap C)</cmath>
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== 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>.
  
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Now, <math>n(A)=\binom{9}{3}+9^3=813</math>. Obviously, <math>n(B)</math> and <math>n(C)</math> are the same. Thus, we have
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== Solution 3==
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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>
  
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==See Also==
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<cmath>2439-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B \cap C)</cmath>
 
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{{incomplete|solution}}
 
  
 
{{AIME box|year=1997|num-b=9|num-a=11}}
 
{{AIME box|year=1997|num-b=9|num-a=11}}
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 18:37, 25 April 2022

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 cards 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 ${3\choose 2}$ ways to pick the two attributes in question. Then there are $3$ ways to pick the value of the first attribute, $3$ ways to pick the value of the second attribute, and $1$ way to arrange the positions of the third attribute, giving us ${3\choose 2} \cdot 3 \cdot 3 = 27$ ways.
  • Case 3: One of the three attributes are the same. There are ${3\choose 1}$ ways to pick the one attribute in question, and then $3$ ways to pick the value of that attribute. Then there are $3!$ ways to arrange the positions of the next two attributes, giving us ${3\choose 1} \cdot 3 \cdot 3! = 54$ ways.
  • Case 4: None of the three attributes are the same. We fix the order of the first attribute, and then there are $3!$ ways to pick the ordering of the second attribute and $3!$ ways to pick the ordering of the third attribute. This gives us $(3!)^2 = 36$ ways.

Adding the cases up, we get $27 + 54 + 36 = \boxed{117}$.

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 $\binom{27}{2} = 27*13 = 351$ possibilities. Note, however, that each set is generated by ${3\choose 2} = 3$ pairs, so we've overcounted by a multiple of 3 and the answer is $\frac{351}{3} = \boxed{117}$.

Solution 3

Treat the sets as ordered. Then for each of the three criterion, there are $3!=6$ choices if the attribute is different and there are $3$ choices is the attribute is the same. Thus all three attributes combine to a total of $(6+3)^3=729$ 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 $3^3=27$ possibilities. Notice that we have counted every set $3!=6$ times by treating the set as ordered. The final solution is then $\frac{729-27}{6}=\boxed{117}$

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

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