Difference between revisions of "2021 AIME II Problems/Problem 6"
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==Solution 1== | ==Solution 1== | ||
− | By PIE, <math>|A|+|B|-|A \cap B| = |A \cup B|</math>, | + | By PIE, <math>|A|+|B|-|A \cap B| = |A \cup B|</math>. Substituting into the equation and factoring, we get that <math>(|A| - |A \cap B|)(|B| - |A \cap B|) = 0</math>, so therefore <math>A \subseteq B</math> or <math>B \subseteq A</math>. WLOG <math>A\subseteq B</math>, then for each element there are <math>3</math> possibilities, either it is in both <math>A</math> and <math>B</math>, it is in <math>B</math> but not <math>A</math>, or it is in neither <math>A</math> nor <math>B</math>. This gives us <math>3^{5}</math> possibilities, and we multiply by <math>2</math> since it could have also been the other way around. Now we need to subtract the overlaps where <math>A=B</math>, and this case has <math>2^{5}=32</math> ways that could happen. It is <math>32</math> because each number could be in the subset or it could not be in the subset. So the final answer is <math>2\cdot 3^5 - 2^5 = \boxed{454}</math>. |
− | ~ math31415926535 | + | ~math31415926535 |
==Solution 2== | ==Solution 2== | ||
Line 112: | Line 112: | ||
The answer is | The answer is | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | &\hspace{5.125mm} \sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] \\ | + | &\hspace{5.125mm}\sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] \\ |
&=\left[2\binom{5}{0}2^{5-0}-\binom{5}{0}\right] + \left[2\binom{5}{1}2^{5-1}-\binom{5}{1}\right] + \left[2\binom{5}{2}2^{5-2}-\binom{5}{2}\right] + \left[2\binom{5}{3}2^{5-3}-\binom{5}{3}\right] + \left[2\binom{5}{4}2^{5-4}-\binom{5}{4}\right] + \left[2\binom{5}{5}2^{5-5}-\binom{5}{5}\right] \\ | &=\left[2\binom{5}{0}2^{5-0}-\binom{5}{0}\right] + \left[2\binom{5}{1}2^{5-1}-\binom{5}{1}\right] + \left[2\binom{5}{2}2^{5-2}-\binom{5}{2}\right] + \left[2\binom{5}{3}2^{5-3}-\binom{5}{3}\right] + \left[2\binom{5}{4}2^{5-4}-\binom{5}{4}\right] + \left[2\binom{5}{5}2^{5-5}-\binom{5}{5}\right] \\ | ||
&=\left[2\left(1\right)2^5-1\right] + \left[2\left(5\right)2^4-5\right] + \left[2\left(10\right)2^3-10\right] + \left[2\left(10\right)2^2-10\right] + \left[2\left(5\right)2^1-5\right] + \left[2\left(1\right)2^0-1\right] \\ | &=\left[2\left(1\right)2^5-1\right] + \left[2\left(5\right)2^4-5\right] + \left[2\left(10\right)2^3-10\right] + \left[2\left(10\right)2^2-10\right] + \left[2\left(5\right)2^1-5\right] + \left[2\left(1\right)2^0-1\right] \\ | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==See | + | ==Solution 4 (Simple Bash)== |
+ | Proceed with Solution 1 to get <math>(|A| - |A \cap B|)(|B| - |A \cap B|) = 0</math>. WLOG, assume <math>|A| = |A \cap B|</math>. Thus, <math>A \subseteq B</math>. | ||
+ | |||
+ | Since <math>A \subseteq B</math>, if <math>|B| = n</math>, there are <math>2^n</math> possible sets <math>A</math>, and there are also <math>{5 \choose n}</math> ways of choosing such <math>B</math>. | ||
+ | |||
+ | Therefore, the number of possible pairs of sets <math>(A, B)</math> is | ||
+ | |||
+ | <cmath> | ||
+ | \sum_{k=0}^{5} 2^n {5 \choose n} | ||
+ | </cmath> | ||
+ | |||
+ | We can compute this manually since it's only from <math>k=0</math> to <math>5</math>, and computing gives us <math>243</math>. We can double this result for <math>B \subseteq A</math>, and we get <math>2(243) = 486</math>. | ||
+ | |||
+ | However, we have double counted the cases where <math>A</math> and <math>B</math> are the same sets. There are <math>32</math> possible such cases, so we subtract <math>32</math> from <math>486</math> to get <math>\boxed{454}</math>. | ||
+ | |||
+ | ~ adam_zheng | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/jEghPVjyHoc | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | ==Video Solution & Set Theory Review== | ||
+ | https://youtu.be/3vcLujj74RM | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | ==See Also== | ||
{{AIME box|year=2021|n=II|num-b=5|num-a=7}} | {{AIME box|year=2021|n=II|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:02, 29 January 2024
Contents
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, . Substituting into the equation and factoring, we get that
, so therefore
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 have 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)
Solution 3 (Principle of Inclusion-Exclusion)
By the Principle of Inclusion-Exclusion (abbreviated as PIE), we have from which we rewrite the given equation as
Rearranging and applying Simon's Favorite Factoring Trick give
from which at least one of the following is true:
Let For each value of
we will use PIE to count the ordered pairs
Suppose There are
ways to choose the elements for
These
elements must also appear in
Next, there are
ways to add any number of the remaining
elements to
(Each element has
options: in
or not in
). There are
ordered pairs for
Similarly, there are
ordered pairs for
To fix the overcount, we subtract the number of ordered pairs that are counted twice, in which There are
such ordered pairs.
Therefore, there are ordered pairs for
Two solutions follow from here:
Solution 3.1 (Binomial Theorem)
The answer is
~MRENTHUSIASM
Solution 3.2 (Bash)
The answer is
~MRENTHUSIASM
Solution 4 (Simple Bash)
Proceed with Solution 1 to get . WLOG, assume
. Thus,
.
Since , if
, there are
possible sets
, and there are also
ways of choosing such
.
Therefore, the number of possible pairs of sets is
We can compute this manually since it's only from to
, and computing gives us
. We can double this result for
, and we get
.
However, we have double counted the cases where and
are the same sets. There are
possible such cases, so we subtract
from
to get
.
~ adam_zheng
Video Solution by Interstigation
~Interstigation
Video Solution & Set Theory Review
~MathProblemSolvingSkills.com
See Also
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.