Example 1-18 The English word "Wellcome" has 3 vowels: e, o, e. e repeats once. 5 consonants: w, $, \mathrm{c}, \mathrm{m}. But one repeats, arranging these 8 letters so that the 3 vowels are not adjacent, how many different arrangements are there?
5 consonants are represented by $\bigcirc$, and their permutations are $5!/ 2=60$. For each permutation, the positions for vowels are inserted as follows: $\bigcirc \cdot \bigcirc \cdot \bigcirc \cdot \bigcirc \cdot \bigcirc \cdot$
For 3 vowels, the first one has 6 choices of positions, i.e., the number of $\cdot$, the second vowel has 5 choices, and the third vowel has 4 choices.
Considering the repetition of $e$, the total number of schemes is
$$
60 \times 6 \times 5 \times 4 \div 2=3600
$$

Alternatively, the 3 vowels can be arranged first. Since $\mathrm{e}$ is repeated, there are 3 schemes. Using $\square$ to represent vowels. For each arrangement
$$
* \square \circledast \square \circledast \square *
$$
$\otimes$ positions must not be empty, and two of the 5 consonants must be chosen to fill these two positions. According to 1 repetition, there should be $4 \div 2=10$ patterns. At this point, the two consonants 1 must be considered separately, discussing the possible permutations where both $\circledast$ are 1; both $\circledast$ are not 1, this pattern has $P(3,2)=3$ kinds, and the remaining 3 consonants have two 1s; one of the two $*$ is 1, this pattern has 6 kinds, and the remaining 3 consonants have one 1. The number of permutations should be calculated separately in each case, which is more complex, left to the reader as an exercise.