Let $A_{n}=\{a_{1}, b_{1}, a_{2}, b_{2}, \ldots, a_{n}, b_{n}\}$ for $n \geq 1$, and consider the 2-configuration $C_{n}$ consisting of $\{a_{i}, b_{i}\}$ for all $1 \leq i \leq n, \{a_{i}, a_{i+1}\}$ for all $1 \leq i \leq n-1$, and $\{b_{i}, b_{i+1}\}$ for all $1 \leq i \leq n-1$. Let $N_{n}$ be the number of subsets of $C_{n}$ that are consistent of order 1 (call these "matchings" of $C_{n}$ ). Consider any matching of $C_{n+2}$. Either $a_{n+2}$ is paired with $b_{n+2}$, in which case the remaining elements of our matching form a matching of $C_{n+1}$; or $a_{n+2}$ is paired with $a_{n+1}$, in which case $b_{n+2}$ must be paired with $b_{n+1}$, and the remaining elements form a matching of $C_{n}$. It follows that $N_{n+2}=N_{n+1}+N_{n}$. By direct calculation, $N_{1}=1$ and $N_{2}=2$, and now computing successive values of $N_{n}$ using the recurrence yields $N_{10}=89$.