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

Latest revision as of 01:29, 14 May 2020

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 after that, to generate $S_{2,2}$ , you take $S_{1,1}$ and add 2 new elements, getting you $\binom{2n-2}{2}$ ways to generate $S_{2,2}$ . 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

Claim: 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 $S_{a+1,b}$ .

Proof: Note $a \leq a+1$ and $b \leq b+1$ , so $S_{a,b}\subseteq S_{a+1,b+1}$ . Let $A$ be a set that contains all the elements in $S_{a+1,b+1}$ that are not in $S_{a,b}$ . $A = S_{a+1,b+1} - S_{a,b}$ . We know $S_{a+1,b+1}$ contains total $a+b+2$ elements. And $S_{a,b}$ contains total $a+b$ elements. That means $A$ contains only 2 elements since $(a+b+2)-(a+b) = 2$ . Let’s call these 2 elements $m, n$ . $A = \{m, n\}$ . $S_{a,b+1}$ contains 1 elements more than $S_{a,b}$ and 1 elements less than $S_{a+1,b+1}$ . That 1 elements has to select from $A$ . It’s easy to see $S_{a,b+1} = S_{a,b}\cup \{m\}$ or $S_{a,b+1} = S_{a,b}\cup \{n\}$ , so there are 2 choice for $S_{a,b+1}$ . Same thing applies to $S_{a+1,b}$ .

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! Fortunately, 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 grid, which gets us $\boxed{(2n)! \cdot 2^{n^2}}$

- AlexLikeMath

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

@@ Line 27: / Line 27: @@
 ===Finishing off===
-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>
+To finish off, we have <math>\frac{(2n)!}{2^n} \cdot 2^{n(n+1)}</math> ways to fill in the grid, which gets us <math>\boxed{(2n)! \cdot 2^{n^2}}</math>
 - AlexLikeMath

2019 USAJMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
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