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Difference between revisions of "2010 AIME I Problems/Problem 7"

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==Problem==
 
 
== Problem ==
 
== Problem ==
 
Define an [[order relation|ordered]] triple <math>(A, B, C)</math> of [[set]]s to be ''minimally intersecting'' if <math>|A \cap B| = |B \cap C| = |C \cap A| = 1</math> and <math>A \cap B \cap C = \emptyset</math>. For example, <math>(\{1,2\},\{2,3\},\{1,3,4\})</math> is a minimally intersecting triple. Let <math>N</math> be the number of minimally intersecting ordered triples of sets for which each set is a subset of <math>\{1,2,3,4,5,6,7\}</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>.
 
Define an [[order relation|ordered]] triple <math>(A, B, C)</math> of [[set]]s to be ''minimally intersecting'' if <math>|A \cap B| = |B \cap C| = |C \cap A| = 1</math> and <math>A \cap B \cap C = \emptyset</math>. For example, <math>(\{1,2\},\{2,3\},\{1,3,4\})</math> is a minimally intersecting triple. Let <math>N</math> be the number of minimally intersecting ordered triples of sets for which each set is a subset of <math>\{1,2,3,4,5,6,7\}</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>.

Latest revision as of 17:03, 9 August 2018

Problem

Define an ordered triple $(A, B, C)$ of sets to be minimally intersecting if $|A \cap B| = |B \cap C| = |C \cap A| = 1$ and $A \cap B \cap C = \emptyset$. For example, $(\{1,2\},\{2,3\},\{1,3,4\})$ is a minimally intersecting triple. Let $N$ be the number of minimally intersecting ordered triples of sets for which each set is a subset of $\{1,2,3,4,5,6,7\}$. Find the remainder when $N$ is divided by $1000$.

Note: $|S|$ represents the number of elements in the set $S$.

Solution

Let each pair of two sets have one element in common. Label the common elements as $x$, $y$, $z$. Set $A$ will have elements $x$ and $y$, set $B$ will have $y$ and $z$, and set $C$ will have $x$ and $z$. There are $7 \cdot 6 \cdot 5 = 210$ ways to choose values of $x$, $y$ and $z$. There are $4$ unpicked numbers, and each number can either go in the first set, second set, third set, or none of them. Since we have $4$ choices for each of $4$ numbers, that gives us $4^4 = 256$.

Finally, $256 \cdot 210 = 53760$, so the answer is $\fbox{760}$.

See Also

2010 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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