Let's number fish sellers with $1, 2 \dots 28$ and fish kinds with $1, 2 \dots 28$. Number the columns and rows of a $28 \times 28$ chart with the numbers of fish sellers and fish kinds, respectively. Let the intersection square of the $i$th row and $j$th column be $(i, j)$ and if we put $B$ into squares $(1, 1), (2, 2) \dots (28, 28)$ and $M$ into the rest, we will end up with the answer $2^{28}-28$ by using induction.

Note that the problem is equivalent to this: Rephrased and Generalized Problem wrote: We are given an $n \times n$ board with every cell being $0$ or $1$. We choose exactly $n$ cells such that all the chosen cells are in different rows and different columns. Then, we create a binary string of length $n$ as follows: For each row, we append the value of the cell that is chosen from that row into our string. For every choice of these $n$ cells, what is the maximum number of different binary string that we can create? Here, $28$ is replaced by $n$ and $k$ is the number of different binary strings that we created. We claim that the answer is $2^n - n$. Let $s_i$ be the complement string of column $i$. (See the figure) It is easy to see that we cannot create the strings $s_1 , s_2 , \dots , s_n$. Thus, $k \leq 2^n - n$. Also, post #2 already gave a construction and that construction is easy to prove by induction as #2 said. Thus we are done.