Let $U=\{0,1\}^{2n}$ and $A_i=\{x\in U: x_i=0 \land x_{i+1}=1\}$ for $1\le i\le 2n-1$. Let $J\in \mathcal{P}_r([2n-1])$ i.e subsets of $[2n-1]$ of cardinality $r$. What is the cardinality of $\cap_{j\in J} A_j$ (solution can use $J$)? and for how many $J\in \mathcal{P}_r([2n-1])$ we have that $\cap_{j\in J} A_j$ isn't empty?
My approach: I tried using exclusion-inclusion in the following way: we know that: $|\cup_{j\in J} A_j|=\sum_{j\in J}|A_j| -\sum_{i<j\in J}|A_i\cap A_j| +...+(-1)^J|\cap_{j\in J} A_j|$ and since $A_i$ freezes $2$ indices we can say that $|A_i|=2^{2n-2}$ for all $i$, and if $j\ne i+1$ then $|A_i\cap A_j|=2^{2n-4}$ and so on. we have $|J|$ elements in the sum of $|A_j|$ and we have $|J| \choose 2$ for the sum of $|A_i\cap A_j|$ and so on. As for $|\cup_{j\in J} A_j|$ that is just $2^{2n}$ because we have all the possible binary strings. perhaps the answer then is:
$|\cap_{j\in J} A_j| = (-1)^J[2^{2n}-|J|2^{2n-2}+$ $|J| \choose 2$ $2^{2n-4}+...+(-1)^{J-1}$ $|J| \choose |J|$ $2^{2n-2|J|}]$ or something similar?
$\bigcap_{j\in J}A_j$ is empty if $j$ and $j+1$ are in $J$ for some $j$. If not, then we just have $|J|=r$ independent constraints on two bits each, so in that case $\left|\bigcap_{j\in J}A_j\right|=2^{2(n-r)}$.
The number of $J\in\mathcal P_r([2n-1])$ for which $\bigcap_{j\in J}A_j$ isn’t empty is the number of binary strings of length $2n-1$ with $r$ ones without consecutive ones. That’s a well-known combinatorial problem for which inclusion–exclusion is a bit of overkill. You have $r$ ones and $k=2n-1-r$ zeros to distribute. Add another zero, glue a zero to the right of each one, distribute the resulting $2n-r$ objects ($2(n-r)$ zeros and $r$ glued pairs of $10$) in $\binom{2n-r}r$ ways and remove the final zero to obtain exactly the desired strings.