11.3. In the language, there are three letters - Sh, U, and Ya. A word is a sequence of 100 letters, exactly 40 of which are vowels (i.e., U or Ya), and the remaining 60 are the letter Sh. What is the maximum number of words that can be chosen so that for any two chosen words, at least one of the hundred positions contains vowels that are different?
Answer. $2^{40}$.

Solution. Example. Consider all $2^{40}$ words where starting from the 41st position, all letters are Ш, and the first $40-\mathrm{U}$ or Я. This set of words satisfies the condition.

Estimate. To each of our $m$ words, we associate $2^{60}$ words by replacing each letter Ш with У or Я (in all possible ways). Note that the resulting $m \cdot 2^{60}$ words consist of the letters У and Я and are pairwise distinct (for words derived from the same word, this is clear from the construction, and for words derived from two different words, it follows from the condition). Thus, $m \cdot 2^{60} \leqslant 2^{100}$ and $m \leqslant 2^{40}$.

Remark. The estimate can be obtained in a different way.

Method 1. Flip a coin 100 times. For each word, consider such an event: for every $i$, if at some position $i$ there is the letter У, then a tail appeared on the $i$-th flip, and if the letter Я, then a head. The probability of such an event is $1 / 2^{40}$, and they are mutually exclusive, so the number of words is no more than $2^{40}$.

Method 2. Suppose more than $2^{40}$ words are chosen. Assign each word a weight of 1. Let the first letter be У in $x$ words and Я in $y$ words, and $x \geqslant y$. Double the weights of all words with the first letter У, and set to zero - with the first letter Я. Next, look at the second letter, and so on. Describe the step of considering the $m$-th letter. Let $p$ be the sum of the weights of words where the $m$-th letter is $\mathrm{Y}$, and $q$ be the sum of the weights of words where the $m$-th letter is Я. If $p \leqslant q$, double the weights of words with the $m$-th letter Я and set to zero - with the $m$-th letter У. Otherwise, do the opposite. As a result of these operations, the total weight does not decrease. After 100 operations, the total weight of all words will be greater than $2^{40}$. In each word, there are only 40 letters У or Я, so the weight of each word is no more than $2^{40}$. Therefore, there will be two words with non-zero weights. Then for them, there will be no position where one has $\mathrm{Y}$ and the other has Я or vice versa, which is a contradiction.