2010 AIME II Problems/Problem 8
Problem
Let be the number of ordered pairs of nonempty sets
and
that have the following properties:
-
,
-
,
- The number of elements of
is not an element of
,
- The number of elements of
is not an element of
.
Find .
Solution
Let us partition the set into
numbers in
and
numbers in
,
Since must be in
and
must be in
(
, we cannot partition into two sets of 6 because
needs to end up somewhere,
or
either).
We have ways of picking the numbers to be in
.
So the answer is .
Note: We have ways of picking the numbers to be in
because there are
numbers in
and since
is already a term in the set we simply have to choose another
numbers from the
numbers that are available.
Solution 2
Regardless of the size of
(ignoring the case when
),
must not be in
and
must be in
.
There are remaining elements whose placements have yet to be determined. Note that the actual value of
does not matter; there is always
necessary element,
forbidden element, and
other elements that need to be distributed. There are
places to put each of these elements, for
possibilities.
However, there is the edge case of is forced not the be in either set, so we must subtract the
cases where
and
have size
.
Thus, our answer is
Solution 3 (PIE and Complementary Counting)
The total number of possible subsets is , which is
. Note that picking a subset from the set leaves the rest of the set to be in the other subset. We exclude
and
since they leave a set empty. We proceed with complementary counting and casework:
We apply the Principle of Inclusion and Exclusion for casework on the complementary cases. We find the ways where the element #elements in is in
, which violates the first condition. Then we find the ways where the elements #elements in
and 12-#elements in
are in set
, which violates only the second condition, and not the first (If we did, we would overcount the union of both cases). This ensures all possible cases are covered.
Case 1: #elements in is an element in
There are =
ways in this case.
Case 2: #elements in and 12-#elements are in
We introduce a subcase where #elements in is not
, since it would also result in
for the other element.
There are
items to choose from
, the
being the
mandatory elements in
. Therefore,
can be from
to
.
There are
ways.
The other subcase is when #elements in
are equal to
. There are
ways for this case, which equals
.
Adding the subcases gives us
.
Adding this with case one gives us . Subtracting this from
gives
.
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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