Find the number of ways in which the letters of the word PERCEPTIVE can be arranged if

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 = $5040 + 25200 = 30240$.