On Problem 5 of IZhO 2015

Tue, 18 Aug 2026 00:00:00 +0000

Problem (IZhO 2015, P5). Let $A_n$ be the set of partitions of the sequence $1,2,\ldots,n$ into several subsequences such that every two neighbouring terms of each subsequence have different parity, and $B_n$ the set of partitions of the sequence $1,2,\ldots,n$ into several subsequences such that all the terms of each subsequence have the same parity (for example, the partition ${(1,4,5,8),(2,3),(6,9),(7)}$ is an element of $A_9$, and the partition ${(1,3,5),(2,4),(6)}$ is an element of $B_6$). Prove that for every positive integer $n$ the sets $A_n$ and $B_{n+1}$ contain the same number of elements.

It might just be me, but the statement seems unnecessarily complicated. On a somewhat funnier note, the notation in the first example looks a little like a set of cycles in the ninth alternating group.

Anyhow, I encourage you to try solving the problem on your own before reading further. The form of the statement points to bijections, and combinatorial bijections are known to be quite varied.

We now give an alternative solution, which incorporates different ideas from the official solution (which I also recommend you read).

Solution.

Since we are asked to prove that two sets have the same number of elements, naturally we shall construct a bijection.

The first useful observation is to think of a partition in $A_n$ as consisting of integers ordered from left to right. We write the entries in every subsequence in increasing order.

Suppose that $1,2,\ldots,k-1$ have already been placed. When we come to $k$, there are only two possibilities:

(a) $k$ begins a new subsequence;

(b) $k$ is placed after the last element of an existing subsequence.

In the second case, the last element of that subsequence necessarily has parity opposite to that of $k$.

We shall build the corresponding partition in $B_{n+1}$ simultaneously through the construction of a tableau.

At the moment just before $k$ is inserted, arrange the blocks we have already constructed in a two-row tableau of the form

$$ \begin{array}{c|c|c|c||c} C_1&C_2&\cdots&C_r&\ast\\ D_1&D_2&\cdots&D_r&G \end{array}. $$

The top row consists of the current subsequences in $A_{k-1}$.

The bottom row consists of the blocks of a partition in $B_k$. Every $C_i$ is paired with the block $D_i$ directly below it, while one block $G$, which we will call distinguished, is left unpaired and placed under the symbol $\ast$.

We maintain the following two rules:

(a) all the elements of $D_i$ have the same parity as the last element of $C_i$;

(b) the distinguished block $G$ contains $k$.

Since $G$ belongs to $B_k$, all its elements consequently have the same parity as $k$.

We begin, before inserting $1$, with

$$ \begin{array}{c} \ast\ {1} \end{array}. $$

There are no blocks in the top row, and ${1}$ is the distinguished block.

We now insert $k$.

Case (a): $k$ begins a new subsequence

Suppose that $k$ starts a new block ${k}$ in the upper row.

We place this new block above the old distinguished block $G$, and introduce the new singleton ${k+1}$ as the distinguished block:

$$ \begin{array}{c||c} &\ast\ &G \end{array} \quad\longmapsto\quad \begin{array}{c||c} {k}&\ast\ G&{k+1} \end{array}. $$

The first rule is preserved, since every element of $G$ has the same parity as $k$. The new distinguished block ${k+1}$ clearly contains $k+1$, so the second rule is preserved as well.

Case (b): $k$ is appended to an existing subsequence

Suppose instead that $k$ is appended to some block $C$.

Let $D$ be the block immediately below $C$ in our tableau. Thus the elements of $D$ have the same parity as the last element of $C$.

Since $k$ may be appended to $C$, the last element of $C$ has parity opposite to that of $k$. But $k+1$ also has parity opposite to that of $k$. Hence

$$ D\cup{k+1} $$

is still monochromatic.

We now perform the following switch:

$$ \begin{array}{c||c} C&\ast\ D&G \end{array} \quad\longmapsto\quad \begin{array}{c||c} C\cup{k}&\ast\ G&D\cup{k+1} \end{array}. $$

In other words, we append $k$ to $C$, move the old distinguished block $G$ underneath it, and move $D$ into the distinguished position after adjoining $k+1$ to it.

The new last element of $C\cup{k}$ is $k$, and $G$ has the same parity as $k$. On the other hand, $D$ has parity opposite to $k$, which is precisely the parity of $k+1$, so $D\cup{k+1}$ is again a valid block of $B_{k+1}$.

All other columns of the tableau are left unchanged.

After inserting $1,2,\ldots,n$, the bottom row is a partition of

$$ {1,2,\ldots,n+1} $$

into monochromatic blocks. In other words, it is an element of $B_{n+1}$.

Thus the procedure defines a map

$$ \Phi:A_n\longrightarrow B_{n+1}. $$

As an example, let us apply the construction to

$$ \pi={(1,4,5,8),(2,3),(6,9),(7)}\in A_9. $$

After inserting $1,2,3,4$, the tableau is

$$ \begin{array}{c|c||c} (1,4)&(2,3)&\ast\ (2,4)&(3)&(1,5) \end{array}. $$

Now $5$ is appended to the first upper block. The block below it is $(2,4)$, while the distinguished block is $(1,5)$. We therefore switch them and append $6$ to the former:

$$ \begin{array}{c||c} (1,4)&\ast\ (2,4)&(1,5) \end{array} \quad\longmapsto\quad \begin{array}{c||c} (1,4,5)&\ast\ (1,5)&(2,4,6) \end{array}. $$

Continuing in the same way eventually gives $$ \Phi(\pi)

{(1,5,9),(2,4,6,10),(3),(7),(8)}, $$ which is indeed an element of $B_{10}$.

It remains to check that the map is a bijection.

Take an arbitrary partition

$$ \sigma\in B_{n+1}. $$

For $1\leq k\leq n+1$, let $\sigma_k$ denote the partition of ${1,\ldots,k}$ obtained by intersecting every block of $\sigma$ with ${1,\ldots,k}$ and deleting the empty intersections.

We reconstruct the upper row successively. Just before reconstructing the insertion of $k$, the bottom row is $\sigma_k$, and its distinguished block is forced: it is the block containing $k$.

Now consider what happens to $k+1$ when we pass from $\sigma_k$ to $\sigma_{k+1}$.

There are again two possibilities.

(a) If $k+1$ is the smallest element of its block in $\sigma$, then it appears in $\sigma_{k+1}$ as a new singleton block ${k+1}$. This can only have arisen from Case (a) above. Hence $k$ begins a new upper block, and the block of $\sigma_k$ containing $k$ is placed underneath it.

(b) Otherwise, $k+1$ joins a block $D$ already present in $\sigma_k$. Since $D$ has the parity of $k+1$, while the distinguished block containing $k$ has the parity of $k$, these are different blocks. By induction, $D$ lies underneath a unique upper block $C$. We therefore append $k$ to $C$ and reverse the switch from Case (b).

Thus at every stage there is exactly one possible inverse operation. Beginning with

$$ \begin{array}{c} \ast\ {1} \end{array} $$

and proceeding through $k=1,2,\ldots,n$, we recover a unique element of $A_n$.

Hence the construction is invertible, so $\Phi$ is a bijection. Therefore

$$ \boxed{|A_n|=|B_{n+1}|}. $$

Bijections | Nikola Veselinov

On Problem 5 of IZhO 2015

Case (a): $k$ begins a new subsequence

Case (b): $k$ is appended to an existing subsequence