求在下列情况下，单词 PERCEPTIVE 中的字母有多少种排列方式：

1. Identify the groups of consonants and vowels, and calculate the number of arrangements for the consonants.

The word PERCEPTIVE consists of $10$ letters in total:

• $6$ consonants (P, P, R, C, T, V) with $2$ identical Ps.
• $4$ vowels (E, E, E, I) with $3$ identical Es.

First, we arrange the $6$ consonants in a row. Since there are $2$ identical Ps, the number of ways to do this is:
$\frac{6!}{2!} = 360$

2. Calculate the number of ways to place and arrange the vowels in the spaces between the consonants so they are separated.

Placing the $6$ consonants creates $7$ available spaces (including the ends) where vowels can be placed:
_ C _ C _ C _ C _ C _ C _

To ensure no two vowels are together, the $4$ vowels must occupy $4$ of these $7$ spaces.
Number of ways to choose $4$ spaces and arrange the vowels (E, E, E, I) into them:
$\binom{7}{4} \times \frac{4!}{3!} = 35 \times 4 = 140$

3. Calculate the total number of valid arrangements using the multiplication principle.

Total number of ways $= 360 \times 140 = 50400$

1. Calculate the number of arrangements where both letters between the two 'P's are 'E's.

Let the two 'P's and the two 'E's between them form a single block: [P E E P].
Since the 'P's and 'E's are identical, there is only $1$ way to form this block.
We arrange this block with the remaining $6$ letters (E, R, C, T, I, V).
This gives $7$ distinct items in total.
Number of arrangements = $1 \times 7! = 5040$.

2. Calculate the number of arrangements where exactly one letter between the two 'P's is an 'E'.

The other letter between the 'P's must be chosen from the $5$ non-'E' letters (R, C, T, I, V).
There are $\binom{5}{1} = 5$ choices for this letter, and it can be arranged with the 'E' inside the block in $2! = 2$ ways.
Number of ways to form the block [P E X P] or [P X E P] = $5 \times 2 = 10$.
We arrange this block with the remaining $6$ letters ($2$ identical 'E's and $4$ distinct non-'E' letters).
This gives $7$ items, with $2$ identical 'E's.
Number of arrangements for these items = $\frac{7!}{2!} = 2520$.
Total arrangements for this case = $10 \times 2520 = 25200$.

3. Sum the arrangements from both cases to find the total number of ways.

Total number of arrangements = $5040 + 25200 = 30240$.