Difference between revisions of "2021 AIME II Problems/Problem 6"
(→Solution 2) |
(→Solution 2) |
||
Line 39: | Line 39: | ||
Next, we analyze each of these cases, separately. | Next, we analyze each of these cases, separately. | ||
− | + | Case 1: <math>|Y| = 0</math> and <math>|Z| \neq 0</math>. | |
In this case, to count the number of solutions, we do the complementary counting. | In this case, to count the number of solutions, we do the complementary counting. | ||
Line 55: | Line 55: | ||
Therefore, following from the complementary counting, the number of solutions in this case is equal to the number of solutions that satisfy <math>|Y| = 0</math> minus the number of solutions that satisfy <math>|Y| = 0</math> and <math>|Z| = 0</math>, i.e., <math>3^5 - 2^5</math>. | Therefore, following from the complementary counting, the number of solutions in this case is equal to the number of solutions that satisfy <math>|Y| = 0</math> minus the number of solutions that satisfy <math>|Y| = 0</math> and <math>|Z| = 0</math>, i.e., <math>3^5 - 2^5</math>. | ||
− | + | Case 2: <math>|Z| = 0</math> and <math>|Y| \neq 0</math>. | |
This case is symmetric to Case 1. Therefore, the number of solutions in this case is the same as the number of solutions in Case 1, i.e., <math>3^5 - 2^5</math>. | This case is symmetric to Case 1. Therefore, the number of solutions in this case is the same as the number of solutions in Case 1, i.e., <math>3^5 - 2^5</math>. | ||
− | + | Case 3: <math>|Y| = 0</math> and <math>|Z| = 0</math>. | |
Recall that this is one part of our analysis in Case 1. Hence, the number solutions in this case is <math>2^5</math>. | Recall that this is one part of our analysis in Case 1. Hence, the number solutions in this case is <math>2^5</math>. |
Revision as of 20:32, 22 March 2021
Problem
For any finite set , let
denote the number of elements in
. FInd the number of ordered pairs
such that
and
are (not necessarily distinct) subsets of
that satisfy
Solution 1
By PIE, , and after some algebra you see that we need
or
. WLOG
, then for each element there are
possibilities, either it is in both
and
, it is in
but not
, or it is in neither
nor
. This gives us
possibilities, and we multiply by
since it could of also been the other way around. Now we need to subtract the overlaps where
, and this case has
ways that could happen. It is
because each number could be in the subset or it could not be in the subset. So the final answer is
.
~ math31415926535
Solution 2
We denote .
We denote
,
,
,
.
Therefore, and the intersection of any two out of sets
,
,
,
is an empty set.
Therefore,
is a partition of
.
Following from our definition of ,
,
, we have
.
Therefore, the equation
can be equivalently written as
This equality can be simplified as
Therefore, we have the following three cases: (1) and
, (2)
and
, (3)
.
Next, we analyze each of these cases, separately.
Case 1: and
.
In this case, to count the number of solutions, we do the complementary counting.
First, we count the number of solutions that satisfy .
Hence, each number in falls into exactly one out of these three sets:
,
,
.
Following from the rule of product, the number of solutions is
.
Second, we count the number of solutions that satisfy and
.
Hence, each number in falls into exactly one out of these two sets:
,
.
Following from the rule of product, the number of solutions is
.
Therefore, following from the complementary counting, the number of solutions in this case is equal to the number of solutions that satisfy minus the number of solutions that satisfy
and
, i.e.,
.
Case 2: and
.
This case is symmetric to Case 1. Therefore, the number of solutions in this case is the same as the number of solutions in Case 1, i.e., .
Case 3: and
.
Recall that this is one part of our analysis in Case 1. Hence, the number solutions in this case is .
By putting all cases together, following from the rule of sum, the total number of solutions is equal to
~ Steven Chen (www.professorchenedu.com)
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.