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Difference between revisions of "2019 USAJMO Problems/Problem 5"

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m (I will add another solution.)
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Proposed by Ricky Liu
 
Proposed by Ricky Liu
  
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==Solution==
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==Solution 1==
 
Note that there are <math>(2n)!</math> ways to choose <math>S_{1, 0}, S_{2, 0}... S_{n, 0}, S_{n, 1}, S_{n, 2}... S{n, n}</math>, because there are <math>2n</math> ways to choose which number <math>S_{1, 0}</math> is, <math>2n-1</math> ways to choose which number to append to make <math>S_{2, 0}</math>, <math>2n-2</math> ways to choose which number to append to make <math>S_{3, 0}</math>... After that, note that <math>S_{n-1, 1}</math> contains the <math>n-1</math> in <math>S_{n-1. 0}</math> and 1 other element chosen from the 2 elements in <math>S_{n, 1}</math> not in <math>S_{n-1, 0}</math> so there are 2 ways for <math>S_{n-1, 1}</math>. By the same logic there are 2 ways for <math>S_{n-1, 2}</math> as well so <math>2^n</math> total ways for all <math>S_{n-1, j}</math>, so doing the same thing <math>n-1</math> more times yields a final answer of <math>(2n)!\cdot 2^{n^2}</math>.
 
Note that there are <math>(2n)!</math> ways to choose <math>S_{1, 0}, S_{2, 0}... S_{n, 0}, S_{n, 1}, S_{n, 2}... S{n, n}</math>, because there are <math>2n</math> ways to choose which number <math>S_{1, 0}</math> is, <math>2n-1</math> ways to choose which number to append to make <math>S_{2, 0}</math>, <math>2n-2</math> ways to choose which number to append to make <math>S_{3, 0}</math>... After that, note that <math>S_{n-1, 1}</math> contains the <math>n-1</math> in <math>S_{n-1. 0}</math> and 1 other element chosen from the 2 elements in <math>S_{n, 1}</math> not in <math>S_{n-1, 0}</math> so there are 2 ways for <math>S_{n-1, 1}</math>. By the same logic there are 2 ways for <math>S_{n-1, 2}</math> as well so <math>2^n</math> total ways for all <math>S_{n-1, j}</math>, so doing the same thing <math>n-1</math> more times yields a final answer of <math>(2n)!\cdot 2^{n^2}</math>.
  
 
-Stormersyle
 
-Stormersyle
 +
 +
==Solution 2==
 +
 +
There are <math>\frac{(2n)!}{2^n}</math> ways to choose <math>S_{0,0}, S_{1,1} .... S_{n,n}</math>, since, there are <math>\binom{2n}{2}</math> ways to choose <math>S_{1,1}</math>, and you can keep going down the line, and you get that there are <math>\frac{(2n)!}{2^n}</math> ways to pick <math>S_{0,0}, S{1,1} ... S{n,n}</math> Then we can fill out the rest of the gird. First, let’s prove a lemma.
 +
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===Lemma===
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If we know what <math>S_{a,b}</math> is and what <math>S_{a+1,b+1}</math> is, then there are 2 choices for both <math>S_{a,b+1}</math> and <math>S_{a+1,b}</math>. It’s easy to see why. Let A be a set that contains all the elements in S_{a+1,b+1} that are not in S_{a,b} By the problem statement, <math>S_{a,b} \subseteq S_{a+1,b}, S_{a,b+1} \subseteq S_{a+1,b+1}</math>, and using the fact that <math> S_{a,b+1}</math> is 1 element larger than <math>S_{a,b}</math> , and 1 element smaller than <math>S_{a+1,b+1}</math> and same with <math>S_{a+1,b}</math>, we get that, <math>S_{a+1,b}, S_{a,b+1} = S_{a,b}+</math> some element, and, <math>S_{a+1,b}, S_{b,a+1} = S_{a+1,b+1}-</math> some other element. We can combine the two equations and get that <math>S_{a,b}+</math>some element <math>= S_{a+1,b+1}-</math>some other element. Rearranging the sides, and main the substitution for A, we get, <math>A =</math> some element + some other element. And, there are obviously 2 ways to pick what the “some element” is, and that leads to 2 choices for both <math>S_{a,b+1}</math> and <math>S_{a+1,b}</math>.
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===Filling in the rest of the grid===
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We used our proved lemma, and we can fill in <math>S_{0,1}, S{1,2} ... S{n-1,n}</math> then we can fill in the next diagonal, until all <math>S_{i,j}</math> are filled, where <math>i \leq j</math> But, we haven’t finished everything! But filling out the rest of the diagonals in a similar fashion is pretty simple. And, it’s easy to see that we have made <math>n(n+1)</math> decisions, each with 2 choices, when filling out the rest of the grid, so there are <math>2^{n
 +
(n+1)}</math> ways to finish off.
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===Finishing off===
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To finish off, we have <math>\frac{(2n)!}{2^n} \cdot 2^{n(n+1)}</math> ways to fill in the gird, which gets us <math>\boxed{(2n)! \cdot 2^{n^2}}</math>
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{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:54, 20 April 2019

Let $n$ be a nonnegative integer. Determine the number of ways that one can choose $(n+1)^2$ sets $S_{i,j}\subseteq\{1,2,\ldots,2n\}$, for integers $i,j$ with $0\leq i,j\leq n$, such that:


1. for all $0\leq i,j\leq n$, the set $S_{i,j}$ has $i+j$ elements; and

2. $S_{i,j}\subseteq S_{k,l}$ whenever $0\leq i\leq k\leq n$ and $0\leq j\leq l\leq n$.


Proposed by Ricky Liu

Solution 1

Note that there are $(2n)!$ ways to choose $S_{1, 0}, S_{2, 0}... S_{n, 0}, S_{n, 1}, S_{n, 2}... S{n, n}$, because there are $2n$ ways to choose which number $S_{1, 0}$ is, $2n-1$ ways to choose which number to append to make $S_{2, 0}$, $2n-2$ ways to choose which number to append to make $S_{3, 0}$... After that, note that $S_{n-1, 1}$ contains the $n-1$ in $S_{n-1. 0}$ and 1 other element chosen from the 2 elements in $S_{n, 1}$ not in $S_{n-1, 0}$ so there are 2 ways for $S_{n-1, 1}$. By the same logic there are 2 ways for $S_{n-1, 2}$ as well so $2^n$ total ways for all $S_{n-1, j}$, so doing the same thing $n-1$ more times yields a final answer of $(2n)!\cdot 2^{n^2}$.

-Stormersyle

Solution 2

There are $\frac{(2n)!}{2^n}$ ways to choose $S_{0,0}, S_{1,1} .... S_{n,n}$, since, there are $\binom{2n}{2}$ ways to choose $S_{1,1}$, and you can keep going down the line, and you get that there are $\frac{(2n)!}{2^n}$ ways to pick $S_{0,0}, S{1,1} ... S{n,n}$ Then we can fill out the rest of the gird. First, let’s prove a lemma.

Lemma

If we know what $S_{a,b}$ is and what $S_{a+1,b+1}$ is, then there are 2 choices for both $S_{a,b+1}$ and $S_{a+1,b}$. It’s easy to see why. Let A be a set that contains all the elements in S_{a+1,b+1} that are not in S_{a,b} By the problem statement, $S_{a,b} \subseteq S_{a+1,b}, S_{a,b+1} \subseteq S_{a+1,b+1}$, and using the fact that $S_{a,b+1}$ is 1 element larger than $S_{a,b}$ , and 1 element smaller than $S_{a+1,b+1}$ and same with $S_{a+1,b}$, we get that, $S_{a+1,b}, S_{a,b+1} = S_{a,b}+$ some element, and, $S_{a+1,b}, S_{b,a+1} = S_{a+1,b+1}-$ some other element. We can combine the two equations and get that $S_{a,b}+$some element $= S_{a+1,b+1}-$some other element. Rearranging the sides, and main the substitution for A, we get, $A =$ some element + some other element. And, there are obviously 2 ways to pick what the “some element” is, and that leads to 2 choices for both $S_{a,b+1}$ and $S_{a+1,b}$.

Filling in the rest of the grid

We used our proved lemma, and we can fill in $S_{0,1}, S{1,2} ... S{n-1,n}$ then we can fill in the next diagonal, until all $S_{i,j}$ are filled, where $i \leq j$ But, we haven’t finished everything! But filling out the rest of the diagonals in a similar fashion is pretty simple. And, it’s easy to see that we have made $n(n+1)$ decisions, each with 2 choices, when filling out the rest of the grid, so there are $2^{n (n+1)}$ ways to finish off.

Finishing off

To finish off, we have $\frac{(2n)!}{2^n} \cdot 2^{n(n+1)}$ ways to fill in the gird, which gets us $\boxed{(2n)! \cdot 2^{n^2}}$


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

See also

2019 USAJMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
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