Difference between revisions of "2022 AIME I Problems/Problem 12"

Revision as of 01:32, 18 June 2022

Problem

For any finite set $X$ , let $| X |$ denote the number of elements in $X$ . Define $S_n = \sum | A \cap B | ,$ where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\left\{ 1 , 2 , 3, \cdots , n \right\}$ with $|A| = |B|$ . For example, $S_2 = 4$ because the sum is taken over the pairs of subsets $(A, B) \in \left\{ (\emptyset, \emptyset) , ( \{1\} , \{1\} ), ( \{1\} , \{2\} ) , ( \{2\} , \{1\} ) , ( \{2\} , \{2\} ) , ( \{1 , 2\} , \{1 , 2\} ) \right\} ,$ giving $S_2 = 0 + 1 + 0 + 0 + 1 + 2 = 4$ . Let $\frac{S_{2022}}{S_{2021}} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find the remainder when $p + q$ is divided by 1000.

Solution 1 (Easy to Understand)

Let's try out for small values of $n$ to get a feel for the problem. When $n=1, S_n$ is obviously $1$ . The problem states that for $n=2, S_n$ is $4$ . Let's try it out for $n=3$ .

Let's perform casework on the number of elements in $A, B$ .

$\textbf{Case 1:} |A| = |B| = 1$

In this case, the only possible equivalencies will be if they are the exact same element, which happens $3$ times.

$\textbf{Case 2:} |A| = |B| = 2$

In this case, if they share both elements, which happens $3$ times, we will get $2$ for each time, and if they share only one element, which also happens $6$ times, we will get $1$ for each time, for a total of $12$ for this case.

$\textbf{Case 3:} |A| = |B| = 3$

In this case, the only possible scenario is that they both are the set $\{1,2,3\}$ , and we have $3$ for this case.

In total, $S_3 = 18$ .

Now notice, the number of intersections by each element $1 \ldots 3$ , or in general, $1 \ldots n$ is equal for each element because of symmetry - each element when $n=3$ adds $6$ to the answer. Notice that $6 = \binom{4}{2}$ - let's prove that $S_n = n \cdot \binom{2n-2}{n-1}$ (note that you can assume this and answer the problem if you're running short on time in the real test).

Let's analyze the element $k$ - to find a general solution, we must count the number of these subsets that $k$ appears in. For $k$ to be in both $A$ and $B$ , we need $A = \{k\} \cup A'| A' \subset \{1,2,\ldots,n\} \land A' \not \subset \{k\}$ and $B = \{k\} \cup B'| B' \subset \{1,2,\ldots,n\} \land B' \not \subset \{k\}$ (Basically, both sets contain $k$ and another subset of $1$ through $n$ not including $k$ ).

For any $0<=l<=n-1$ that is the size of both $A'$ and $B'$ , the number of ways to choose the subsets $A'$ and $B'$ is $\binom{n-1}{l}$ for both subsets, so the total number of ways to choose the subsets are $\binom{n-1}{l}^2$ . Now we sum this over all possible $l$ 's to find the total number of ways to form sets $A$ and $B$ that contain $k$ . This is equal to $\sum_{l=0}^{n-1} \binom{n-1}{l}^2$ . This is a simplification of Vandermonde's identity, which states that $\sum_{k=0}^{r} \binom{m}{k} \cdot \binom{n}{r-k} = \binom{m+n}{r}$ . Here, $m$ , $n$ and $r$ are all $n-1$ , so this sum is equal to $\binom{2n-2}{n-1}$ . Finally, since we are iterating over all $k$ 's for $n$ values of $k$ , we have $S_n = n \cdot \binom{2n-2}{n-1}$ , proving our claim.

We now plug in $S_n$ to the expression we want to find. This turns out to be $\frac{2022 \cdot \binom{4042}{2021}}{2021 \cdot \binom{4040}{2020}}$ . Expanding produces $\frac{2022 \cdot 4042!\cdot 2020! \cdot 2020!}{2021 \cdot 4040! \cdot 2021! \cdot 2021!}$ .

After cancellation, we have $\frac{2022 \cdot 4042 \cdot 4041}{2021 \cdot 2021 \cdot 2021} \implies \frac{4044\cdot 4041}{2021 \cdot 2021}$

$4044$ and $4041$ don't have any common factors with $2021$ , so we're done with the simplification. We want to find $4044 \cdot 4041 + 2021^2 \pmod{1000} = 44 \cdot 41 + 21^2 \pmod{1000} = 1804+441 \pmod{1000} = 2245 \pmod{1000} = \boxed{245}$

~KingRavi ~Edited by MY-2

Linearity of Expectation Solution

We take cases based on the number of values in each of the subsets in the pair. Suppose we have $k$ elements in each of the subsets in a pair (for a total of n elements in the set). The expected number of elements in any random pair will be $n \cdot \frac{k}{n} \cdot \frac{k}{n}$ by linearity of expectation because for each of the $n$ elements, there is a $\frac{k}{n}$ probability that the element will be chosen. To find the sum over all such values, we multiply this quantity by $\binom{n}{k}^2$ . Summing, we get $\sum_{k=1}^{n} \frac{k^2}{n} \binom{n}{k}^2$ Notice that we can rewrite this as $\sum_{k=1}^{n} \frac{1}{n} \left(\frac{k \cdot n!}{(k)!(n - k)!}\right)^2 = \sum_{k=1}^{n} \frac{1}{n} n^2 \left(\frac{(n-1)!}{(k - 1)!(n - k)!}\right)^2 = n \sum_{k=1}^{n} \binom{n - 1}{k - 1}^2 = n \sum_{k=1}^{n} \binom{n - 1}{k - 1}\binom{n - 1}{n - k}$ We can simplify this using Vandermonde's identity to get $n \binom{2n - 2}{n - 1}$ . Evaluating this for $2022$ and $2021$ gives $\frac{2022\binom{4042}{2021}}{2021\binom{4040}{2020}} = \frac{2022 \cdot 4042 \cdot 4041}{2021^3} = \frac{2022 \cdot 2 \cdot 4041}{2021^2}$ Evaluating the numerators and denominators mod $1000$ gives $804 + 441 = 1\boxed{245}$

- pi_is_3.14

Solution 2 (Rigorous)

For each element $i$ , denote $x_i = \left( x_{i, A}, x_{i, B} \right) \in \left\{ 0 , 1 \right\}^2$ , where $x_{i, A} = \Bbb I \left\{ i \in A \right\}$ (resp. $x_{i, B} = \Bbb I \left\{ i \in B \right\}$ ).

Denote $\Omega = \left\{ (x_1, \cdots , x_n): \sum_{i = 1}^n x_{i, A} = \sum_{i = 1}^n x_{i, B} \right\}$ .

Denote $\Omega_{-j} = \left\{ (x_1, \cdots , x_{j-1} , x_{j+1} , \cdots , x_n): \sum_{i \neq j} x_{i, A} = \sum_{i \neq j} x_{i, B} \right\}$ .

Hence, $\begin{align*} S_n & = \sum_{(x_1, \cdots , x_n) \in \Omega} \sum_{i = 1}^n \Bbb I \left\{ x_{i, A} = x_{i, B} = 1 \right\} \\ & = \sum_{i = 1}^n \sum_{(x_1, \cdots , x_n) \in \Omega} \Bbb I \left\{ x_{i, A} = x_{i, B} = 1 \right\} \\ & = \sum_{i = 1}^n \sum_{(x_1, \cdots , x_{i-1} , x_{i+1} , \cdots , x_n) \in \Omega_{-i}} 1 \\ & = \sum_{i = 1}^n \sum_{j=0}^{n-1} \left( \binom{n-1}{j} \right)^2 \\ & = n \sum_{j=0}^{n-1} \left( \binom{n-1}{j} \right)^2 \\ & = n \sum_{j=0}^{n-1} \binom{n-1}{j} \binom{n-1}{n-1-j} \\ & = n \binom{2n-2}{n-1} . \end{align*}$

Therefore, $\begin{align*} \frac{S_{2022}}{S_{2021}} & = \frac{2022 \binom{4042}{2021}}{2021 \binom{4040}{2020}} \\ & = \frac{4044 \cdot 4041}{2021^2} . \end{align*}$

This is in the lowest term. Therefore, modulo 1000, $\begin{align*} p + q & \equiv 4044 \cdot 4041 + 2021^2 \\ & \equiv 44 \cdot 41 + 21^2 \\ & \equiv \boxed{\textbf{(245) }} . \end{align*}$

~Steven Chen (www.professorchenedu.com)

Revision as of 21:11, 26 May 2022 (view source) My-2 (talk \| contribs) m (→‎Solution 1 (Easy to Understand)) ← Older edit		Revision as of 01:32, 18 June 2022 (view source) Oceanair (talk \| contribs) m Newer edit →
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	After cancellation, we have <cmath>\frac{2022 \cdot 4042 \cdot 4041}{2021 \cdot 2021 \cdot 2021} \implies \frac{4044\cdot 4041}{2021 \cdot 2021}</cmath>		After cancellation, we have <cmath>\frac{2022 \cdot 4042 \cdot 4041}{2021 \cdot 2021 \cdot 2021} \implies \frac{4044\cdot 4041}{2021 \cdot 2021}</cmath>

−	<math>4044</math> and <math>4041</math> don't have any common factors with <math>2021</math>, so we're done with the simplification. We want to find <math>4044 \cdot 4041 + 2021^2 \pmod{1000} = 44 \cdot 41 + 21^2 \pmod{1000} = 1804+~~481~~ \pmod{1000} = 2245 \pmod{1000} = \boxed{245}</math>	+	<math>4044</math> and <math>4041</math> don't have any common factors with <math>2021</math>, so we're done with the simplification. We want to find <math>4044 \cdot 4041 + 2021^2 \pmod{1000} = 44 \cdot 41 + 21^2 \pmod{1000} = 1804+441 \pmod{1000} = 2245 \pmod{1000} = \boxed{245}</math>

2022 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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