Great problem! Sketch of the official solution: Let $M$ be the set of the set of the chosen words, and suppose it works (every two words have a position, where the one has $B$, and the other has $C$) We construct function $f$ from $M$ to $S$, where $S$ is the set of the words, which have only $B$ and $C$ and have length $100$ (obviously they are $2^{100}$). For each word in $M$, replace each $A$ with $B$ or $C$ and we assign it the $2^{60}$ new words. We now have assigned $2^{60}.|M|$ words of only $B$ and $C$ to the words in $M$, and note that these new words are distinct (if there are two identical, there is no position in which the one has $B$, and the other has $C$). So $2^{60}.|M|<=2^{100}$, which proves the bound $|M|<=2^{40}$. The equality case is to take each of the chosen words to have its $60$ last positions to be $A$, and the first $40$ are only $B$ and $C$ and the first parts (the first $40$ positions) are all distinct (thus we have chosen exactly $2^{40}$).

Our answer is $2^{40}$. We can get a set of such words by setting the first $60$ letters to $A$ and varying the remaining letters between $B$ and $C$. Any two such words must differ in at least one of the last $40$ positions else they are the same word, and if they do differ, they must consist of $\{B, C\}$, which are two consonants. Next we show that the set of words picked, call it $S$, has size $|S| \leq 2^{40}$. Indeed, consider the multiset $S'$ formed by: for each word $s$ in $S$, consider each of the $2^{60}$ words formed by replacing each of the $60$ instances of $A$ in $s$ with either $B$ or $C$, and slap them into $S'$. Clearly, $|S'| = 2^{60}|S|$. Note that in fact, no two strings in $S'$ can be the same $-$ if some two $s_1, s_2 \in S'$ were identical, then no matter how we replace some $60$ letters in each of $s_1, s_2$ with $A$, there still cannot be a position where $s_1$ and $s_2$ have letters $B$ and $C$ (in some order). Thus, $|S'| \leq 2^{100}$, as there are $2^{100}$ total strings whose letters are either $B$ or $C$. This proves our bound\[2^{60}|S| = |S'| \leq 2^{100} \implies |S| \leq 2^{40}\]and we are done.

Solved with Elliott Liu and Ryan Yang. The answer is \(2^{40}\), achieved by taking all words whose first 60 characters are \(A\) and whose last 40 characters are either \(B\) and \(C\). Now we show this is optimal. Suppose we select a set \(S\) of words such that any two words have differing consonants in at least one position. For each string \(s\in S\), consider the \(2^{60}\) strings obtained by replacing each \(A\) in \(s\) with either \(B\) or \(C\). It can be seen that all these strings are distinct, and there are at most \(2^{100}\) strings total, so \(S\) has size at most \(2^{40}\).

We claim, that the answer is $2^{40}.$ Example: pick all words with $A$ on first $60$ positions and either $B$ or $C$ on each other. Estimation. Consider any set $\mathcal{P}$ of words, which satisfy the condition, and for it's every word consider $2^{60}$ sequences of letters, obtained by replacing all $A$ with either $B$ or $C$; it's obvious that all such sequences are different, and let $\mathcal{Q}$ denotes set of them: $$|\mathcal{P}|=|\mathcal{Q}|:2^{60}\leq 2^{100}:2^{60}=2^{40}.$$

Amazing problerm

Beautiful The answer is $2^{40}$. A construction is to pick $2^{40}$ distinct sequences of $B$ and $C$ which are $40$ letters long and then add $60$ copies of $A$ to the end. We now prove the bound where $60$ is replaced by $k \geq 0$ by induction on $k$, with the $k=0$ case being obvious. Now consider the scenario with $k+1$ letters which are $A$. Partition them into sets $A$, $B$, and $C$ based on their ending letter. Let $B_B$ be the set formed by switching the rightmost $A$ of every word in $B$ to $B$ and $B_C$ formed by switching that letter to $C$, and define $C_B$ and $C_C$ similarly. Then chop off the rightmost letter of these four sets as well as the set $A$, forming $A'$, etc. It is easy to see that if some consonants differ in some position between $A$ and $B$ then they still differ between $A'$, $B_B'$, and $B_C'$, and the same is true for $A'$, $C_B'$, and $C_C'$. Thus every pair of (distinct) elements in the multiset $|A' \cup B_B' \cup B_C'|$ differ at some consonant, so by hypothesis $$|A'|+|B_B'|+|B_C'|=|A|+2|B| \leq 2^{40}$$and likewise $|A|+2|C| \leq 2^{40}$. Summing and dividing by $2$, we get $|A \cup B \cup C|=|A|+|B|+|C| \leq 2^{40}$, as desired. $\blacksquare$ Remark: The induction is not necessary at all but I approached the problem from that angle, and it doesn't make the problem harder either (this has happened with a few other combinatorics problems where my solution is the same as everyone else's but I have induction for some reason). The inductive approach was motivated by the following consideration: after playing around a bit it seems like the answer is $2^{40}$ regardless of how many letters are $A$, and the bound is obvious if none of the letters are $A$. It seems slightly easier to work with the case, for instance, where $1$ of the letters is $A$, and after playing around a bit more we find the idea of performing the partition and then switching the letter $A$ in the sets $B$ and $C$ to the letters $B$ and $C$ (personally, at the very first I only had the idea to swap the letter $A$ in set $B$ to letter $B$, but from there it's a quick adaptation), and this can then be made to work for a general inductive step.

Here is a solution using the Probabilistic Method (motivated by the fact that a very similar problem was encountered in 18.226): We reformulate the problem as follows: reformulation wrote: Consider some subsets $A_1, B_1, \ldots A_k, B_k$ of $[100]$, where for all $i$, $|A_i| + |B_i| = 40$, $A_i \cap B_i = \emptyset$, and for every $i \neq j$, at least one of $A_i \cap B_j$ and $A_j \cap B_i$ is nonempty. Find the maximum value of $k$. The proof now proceeds as follows. Choose independently and uniformly at random a real from $[0, 2]$ for each element in $[100]$. Call $i$ nice if the labels of all the elements in $A_i$ are $< 1$ and the labels of all the elements in $B_i$ are $\geq 1$. Clearly every $i$ has a probability of $2^{-40}$ of being nice, so the expected number of nice $i$s is $k2^{-40}$. But the condition that at least one of $A_i \cap B_j$, $A_j \cap B_i$ is nonempty guarantees that at most one $i$ can be nice, so thus $k2^{-40} \leq 1$, and therefore $k \leq 2^{40}$. Equality is given when all the words have $A$ as their last $60$ letters.