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Revision as of 12:21, 4 May 2020
Contents
Problem 20
Let be the set
. For
, define
to mean that either
or
. How many ordered triples
of elements of
have the property that
,
, and
?
Solution 1
Imagine that the 19 numbers are just 19 persons sitting evenly around a circle ; each of them is facing to the center.
One may check that if and only if
is one of the 9 persons on the left of
, and
if and only if
is one of the 9 persons on the right of
. Therefore, "
and
and
" implies that
cuts the circumference of
into three arcs, each of which has no more than
numbers sitting on it (inclusive).
We count the complement: where the cut generated by has ONE arc that has more than
persons sitting on. Note that there can only be one such arc because there are only
persons in total.
Suppose the number of persons on the longest arc is . Then two places of
are just chosen from the two end-points of the arc, and there are
possible places for the third person. Once the three places of
are chosen, there are three possible ways to put
into them clockwise. Also, note that for any
, there are
ways to choose an arc of length
. Therefore the total number of ways (of the complement) is
So the answer is
NOTE: this multiple-choice problem can be done even faster -- after we realized the fact that each choice of the three places of corresponds to
possible ways to put them in, and that each arc of length
has
equitable positions, it is evident that the answer should be divisible by
, which can only be
from the five choices.
Solution 2
First, we can find out that the only that satisfy the conditions in the problem are
,
, and
.
Consider the 1st set of conditions for . We get that there are
cases for the first set of conditions.
Since the 2nd and 3rd set of conditions are simply rotations of the 1st set, the total number of cases is
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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