Difference between revisions of "2010 AIME I Problems/Problem 7"
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Finally, <math>256 \cdot 210 = 53760</math>, so the answer is <math>\fbox{760}</math>. | Finally, <math>256 \cdot 210 = 53760</math>, so the answer is <math>\fbox{760}</math>. | ||
− | == See | + | == See Also == |
{{AIME box|year=2010|num-b=6|num-a=8|n=I}} | {{AIME box|year=2010|num-b=6|num-a=8|n=I}} | ||
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 17:03, 9 August 2018
Problem
Define an ordered triple of sets to be minimally intersecting if
and
. For example,
is a minimally intersecting triple. Let
be the number of minimally intersecting ordered triples of sets for which each set is a subset of
. Find the remainder when
is divided by
.
Note: represents the number of elements in the set
.
Solution
Let each pair of two sets have one element in common. Label the common elements as ,
,
. Set
will have elements
and
, set
will have
and
, and set
will have
and
. There are
ways to choose values of
,
and
. There are
unpicked numbers, and each number can either go in the first set, second set, third set, or none of them. Since we have
choices for each of
numbers, that gives us
.
Finally, , so the answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.