Let us generalize the problem and formulate it in graph theory language. Generalized problem wrote: Let $n,m$ be positive integers. Let $U,V$ be sets where $|U|=n$ and $|V|=m$. In a bipartite graph $G=(U,V,E)$ we say that a subset $S\subset U$ is dominant if $|N_G(S)|\ge|S|$. Show that for every integer $k$ with $0\le k\le\sum_{i=1}^m\binom{n}{i}$, there exists an arrangement with exactly $k$ nonempty dominant sets. Note that plugging $m=n$ implies the original statement, as $\sum_{i=1}^n\binom{n}{i}=2^n-1$. We prove the statement by induction on $n$. The base case $n=1$ is obvious. Now assume $n>1$. If $k=0$, then choose the empty graph on $A,B$. So now assume $k\ge 1$. Let $A_1,\ldots,A_n$ be the vertices of $A$ and let $B_1,\ldots,B_m$ be the vertices of $B$. Moreover let $t$ be the maximal integer such that $t\in\{1,2,\ldots,m\}$ and $\sum_{i=0}^{t-1}\binom{n-1}{i}\le k$. Note that $t$ indeed exists as $\binom{n-1}{0}=1\le k$. In the constructed graph we will connect $A$ only to $B_1,\ldots,B_t$. In addition, we connect $A_1$ to all $B_1,\ldots,B_t$. Hence the dominant sets containing $A_1$ are exactly the number of subsets of $A$ containing $A_1$ with size at most $t$. This number is exactly $\sum_{i=0}^{t-1}\binom{n-1}{i}$. Therefore, to use induction hypothesis, it suffices to prove that $$0\le k-\sum_{i=0}^{t-1}\binom{n-1}{i}\le\sum_{i=1}^t\binom{n-1}{i}.$$The left-hand side is true by assumption. For the right-hand side, we distinguish between two cases. Case 1. $t=m$. By Pascal's identity, \begin{align*} k-\sum_{i=0}^{m-1}\binom{n-1}{i}&\le\sum_{i=1}^m\binom{n}{i}-\sum_{i=0}^{m-1}\binom{n-1}{i}\\ &=\sum_{i=1}^m\left[\binom{n}{i}-\binom{n-1}{i-1}\right]\\ &=\sum_{i=1}^m\binom{n-1}{i}. \end{align*} Case 2. $t<m$. By maximality of $k$, \begin{align*} k-\sum_{i=0}^{t-1}\binom{n-1}{i}&\le k-\binom{n-1}{0}\\ &\le \sum_{i=0}^t\binom{n-1}{i}-\binom{n-1}{0}\\ &=\sum_{i=1}^t\binom{n-1}{i}. \end{align*} This ends our proof.