Problem (Combinatorics):
Given an arbitrary finite sequence of letters (represented as a word), a subsequence is a sequence of one or more letters that appear in the same order as in the original sequence. For example, $N, C T, O T T$, and CONTEST are subsequences of the word CONTEST, but NOT, ONSET, and TESS are not. Assuming the standard English alphabet $\{A, B, \ldots, Z\}$, compute the number of distinct four-letter "words" for which $E E$ is a subsequence.

Solution:
Divide into cases according to the number of $E$ 's in the word. If there are only two $E$ 's, then the word must have two non- $E$ letters, represented by ?'s. There are $\left(\begin{array}{l}4 \\ 2\end{array}\right)=6$ arrangements of two $E$ 's and two ?'s, and each of the ?'s can be any of 25 letters, so there are $6 \cdot 25^{2}=3750$ possible words. If there are three $E$ 's, then the word has exactly one non- $E$ letter, and so there are 4 arrangements times 25 choices for the letter, or 100 possible words. There is one word with four $E$ 's, hence a total of 3851 words.

Final answer: 3851