Recursive formula for number of $n\times m\times p$ cubes of $0$'s and $1$'s with a special property. If $L(n, m)$ denotes the number of $n\times m$ matrices consisting only of $0$'s and $1$'s, such that there is no column or row consisting only of $0$'s, we get a nice recursive formula which enables us to compute them with ease: $$\sum_{j=1}^m \binom{m}{j} L(n, j) = (2^m-1)^n.$$
Now, let's consider two kinds of numbers, $L_1(n, m, p)$ and $L_2(n, m, p)$. The first kind of numbers will be the amount of $n\times m\times p$ cubes consisting only of $0$'s and $1$'s such that if we cross the cube with a line perpendicular to one of the axis, we get a non-zero vector. $L_2(n, m, p)$ is defined the same, expect that instead of taking lines, we take planes orthogonal to one of the axis. Clearly $L_1(n, m, p)\leq L_2(n, m, p)$.
Can we get a similar nice looking recursive formula for these numbers? Perhaps in terms of $L(n, m)$?
To be more precise, let me try to explain what exactly do I mean with $L_1(n, m, p)$. For example, you take some $1\leq i\leq n$, $1\leq j\leq m$ then for the cube $B$ we need to have $B_{ijk} = 1$ for some $1\leq k\leq p$.
Similarly, with $L_2(m, n, p)$ we can take some $1\leq i\leq n$, then there must be $1\leq j\leq m$, $1\leq k\leq p$ such that $B_{ijk} = 1$.
 A: One approach is classic inclusion-exclusion.
The argument is much simpler for $L_2(n,m,p)$. I'll use $1 \leq i \leq n$, $1 \leq j \leq m$, $1 \leq k \leq p$ throughout.

*

*Let $P_i$ be the set of such matrices with $B_{i,j,k} = 0$ for all $j,k$.

*Let $Q_j$ be the set of such matrices with $B_{i,j,k} = 0$ for all $i,k$.

*Let $R_k$ be the set of such matrices with $B_{i,j,k} = 0$ for all $i,j$.

$P_i$ forces $mp$ entries to be $0$, so $|P_i| = 2^{nmp - mp}$. Similarly an $a$-fold intersection of $P_i$'s does not force $(n-a)mp$ entries to be $0$, so the result has size $2^{(n-a)mp}$. Likewise, an $a$-fold intersection of $P_i$'s intersected with a $b$-fold intersection of $Q_j$'s and a $c$-fold intersection of $R_k$'s does not force $(n-a)(m-b)(p-c)$ entries to be $0$, so has size $2^{(n-a)(m-b)(p-c)}$.
By the principle of inclusion-exclusion, the number of such matrices in none of these sets is
$$L_2(n,m,p) = \sum_{a=0}^n \sum_{b=0}^m \sum_{c=0}^p (-1)^{a+b+c} \binom{n}{a} \binom{m}{b} \binom{p}{c} 2^{(n-a)(m-b)(p-c)}. \qquad(*)$$
Analogous reasoning gives
$$L(n,m) = \sum_{a=0}^n \sum_{b=0}^m (-1)^{a+b} \binom{n}{a} \binom{m}{b} 2^{(n-a)(m-b)}. \qquad(**)$$
You can go from (**) to your formula by repeated applications of the binomial theorem:
$$\begin{align*}
  \sum_{m=0}^M \binom{M}{m} L(N,m)
    &= \sum_{m=0}^M \binom{M}{m} \sum_{a=0}^N \sum_{b=0}^m (-1)^{a+b} \binom{N}{a} \binom{m}{b} 2^{(N-a)(m-b)} \\
    &= \sum_{a=0}^N (-1)^a \binom{N}{a} \sum_{m=0}^M \binom{M}{m} (2^{N-a}-1)^m \\
    &= \sum_{a=0}^N (-1)^a \binom{N}{a} (2^{N-a})^M \\
    &= (2^M-1)^N.
\end{align*}$$
Similarly from (*) we get
$$\sum_{m=0}^M \sum_{p=0}^P \binom{M}{m} \binom{P}{p} L_2(N,m,p) = (2^{MP}-1)^N$$
since the left-hand side is
$$\begin{align*}
  &\sum_{a=0}^N (-1)^a \binom{N}{a} \sum_{m=0}^M \binom{M}{m} \sum_{b=0}^m (-1)^b \binom{m}{b} \sum_{p=0}^P \binom{P}{c} \sum_{c=0}^p (-1)^c \binom{p}{c} 2^{(N-a)(m-b)(p-c)} \\
  &= \sum_{a=0}^N (-1)^a \binom{N}{a} \sum_{m=0}^M \binom{M}{m} \sum_{b=0}^m (-1)^b \binom{m}{b} \sum_{p=0}^P \binom{P}{c} (2^{(N-a)(m-b)}-1)^p \\
  &= \sum_{a=0}^N (-1)^a \binom{N}{a} \sum_{m=0}^M \binom{M}{m} \sum_{b=0}^m (-1)^b \binom{m}{b} (2^{P(N-m)})^{m-b} \\
  &= \sum_{a=0}^N (-1)^a \binom{N}{a} \sum_{m=0}^M \binom{M}{m} (2^{P(N-a)}-1)^m \\
  &= \sum_{a=0}^N (-1)^a \binom{N}{a} (2^{MP})^{N-a} \\
  &= (2^{MP}-1)^N.
\end{align*}$$
(Obviously this will all generalize to hypercubes.)
You can get a formula for $L_1(m, p, n)$ along these lines, but the multi-fold intersections of the analogues of the $P$'s, $Q$'s, and $R$'s are significantly more annoying, since lines can either be skew or intersect. You'd need to index the intersections by subsets of $(i, j)$'s, $(j, k)$'s, and $(k, i)$'s, tracking how many times these collections themselves intersect. One would expect a recursive formula to exist, but it may not be transparent how to get one from this approach. Perhaps I'm overly pessimistic, but I see no reason to expect there to be a "nice" answer for $L_1(n, m, p)$.
