The formula expressing, the number of all sequences $(X_1, . . . , X_m)$ of length $m$ of subsets of the set $[n] = \{ 1, . . . , n \}$

Question

Find the formula (may be in the form of a sum) expressing, for fixed natural numbers $m, n \geq 1$, the number of all sequences $(X_1, . . . , X_m)$ of length $m$ of subsets of the set $[n] = \{ 1, . . . , n \}$ such that:

$$\bigcup^m_{i=1} X_i = [n]$$

and each of the sets $X_1, . . . , X_m$ has an odd number of elements.

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We need to put every element of set $n$ in one or more of $m$ buckets - that way the sum of $m$ buckets will give us the whole set $n$. Then we exchange elements between buckets since buckets are distinguishable.

Thinking that way we can enumerate through all possible numbers of buckets for every element of the set $[n] = \{ 1, . . . , n \}$ (getting the sum) and exchange elements between chosen numbers of buckets. We can repeat that for every element of the set $[n] = \{ 1, . . . , n \}$ (getting the power n).

$$\big( \sum_{i=1}^m {{m}\choose{i}} i! \big)^n$$

The problem is that counting that way, I don't see any way to control if each of sets $X_1, . . . , X_m$ has an odd number of elements.

Any help would be appreciated.

Answer 1

Here is a method which does not work, but which comes close to working. Afterwards, I will discuss how to fix it:

1. First, choose a list $(Y_1,\dots,Y_{m})$, where $Y_i$ are arbitrary subsets of $[n-1]$, such that $Y_1\cup \dots \cup Y_{m}=[n-1]$. The number of ways to do this is $$(2^m-1)^{n-1}.$$To see this, for each $i\in [n-1]$, let $S_i=\{s\in [m]\mid i\in X_s\}$. Note that $S_i$ must be a nonempty subset of $[m]$. There are $2^m-1$ ways to choose a nonempty subset of $[m]$, and this must be done for every element of $[n-1]$.

2. For each $i\in \{1,\dots,n\}$, define $$ X_i= \begin{cases} Y_i\cup \{n\} & \text{if }|Y_i|\text{ is even} \\ Y_i & \text{if $|Y_i|$ is odd}. \end{cases} $$Note that $|X_i|$ is odd for each $i\in [m]$. Furthermore, as long as at least one of the $Y_i$'s was even, then it will be the case that $\bigcup_{i=1}^m X_i=[n]$.

So, when we follow the above procedure, we get $(2^m-1)^{n-1}$ lists of the form $(X_1,\dots,X_m)$, where each $|X_i|$ is odd. Most of these lists have the property that the union is $[n]$, but there are some extra "bad" lists where we only have $\bigcup_{i=1}^m X_i=[n-1]$, because all of the $|Y_i|$ were odd. It follows that the final answer is $$ (2^m-1)^{n-1}-(\text{# bad lists})\tag1 $$ There is one bad list for every way to choose $(Y_1,\dots,Y_m)$ such that every $Y_i$ is odd, and $\bigcup_{i=1}^m Y_i=[n-1]$. We can now see that counting bad lists is just a smaller version of the original problem, with $n-1$ instead of $n$. This leads to a solution by induction. Namely, if we let $f(m,n)$ be the solution to this problem, then $(1)$ becomes $$ f(m,n)=(2^m-1)^{n-1}-f(m,n-1)\tag2 $$ Using $(2)$, you can prove the following fact by induction: $$ f(m,n)=(2^m-1)^{n-1}-(2^m-1)^{n-2}+(2^m-1)^{n-3}+\dots+(-1)^{n-2}(2^m-1)^1+(-1)^{n-1} $$ This formula is a finite geometric series, so it admits a closed form: $$ f(m,n)=\frac{(2^m-1)^n-(-1)^n}{2^m} $$