1997 AIME Problems/Problem 10

Revision as of 03:20, 7 November 2021 by Arrow-arrow (talk | contribs) (Solution 1)

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 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 (ProblemsAnswer KeyResources)
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. AMC logo.png