From 8 English letters $A, B, C, D, E, X, Y, Z$, any 5 letters (letters can be repeated) form a “word”. All possible “words” are arranged in “dictionary order” (i.e., the order in which English words are arranged in an English-Chinese dictionary), resulting in a “word list”:

Try to find the number of “words” located between the “word” $C Z Y E B$ and the “word” $X C E D A$ (excluding these two words).

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Number Theory, Bases, and Position.
We will map 8 English letters $A, B, C, D, E, X, Y, Z$ to the numbers $0, 1, 2, 3, 4, 5, 6, 7$ respectively; thus, the sequence of "words" $A A A A A, A A A A B, \ldots, A A A A Z, A A A B A, A A A B B, \ldots, Z Z Z Z Y, Z Z Z Z Z$ is the same as the sequence of five-digit octal numbers $00000,00001, \ldots, 00007,00010,00011, \ldots, 77776,77777$. Therefore, the number of "words" between "word" CZYEB and "word" XCEDA (excluding these two words) is the same as the number of five-digit octal numbers between 27641 and 52430 (excluding these two numbers).
$$
\begin{array}{l}
(27641)_{8}=2 \times 8^{4}+7 \times 8^{3}+6 \times 8^{2}+4 \times 8^{1}+1 \times 8^{0}=2 \times 4096+7 \times 512+6 \times 64+4 \times 8+1 \times 1=(12193)_{10} ; \\
(52430)_{8}=5 \times 8^{4}+2 \times 8^{3}+4 \times 8^{2}+3 \times 8^{1}+0 \times 8^{0}=5 \times 4096+2 \times 512+4 \times 64+3 \times 8+0 \times 1=(21784)_{10} ;
\end{array}
$$

There are $21783-12194+1=9590$ numbers between 12193 and 21784 (excluding these two numbers); hence, the number of "words" between "word" CZYEB and "word" XCEDA (excluding these two words) is 9590.