2019 USAJMO Problems/Problem 5

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

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 $S_{a,b}$ . $A$ contains 2 elements, since $(a+1+b+1)-(a+b) = 2$ . Let’s call the 2 elements, $m,n$ . It’s easy to see $S_{a,b+1} = S_{a,b}\cup$ {m} or {n}, so there are 2 possibilities 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 gird, 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.

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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