Take the task of arranging the letters and break it into stages.

Stage 1: "glue" the three vowels (A, A and E) together.
There are three possible outcomes: AAE, AEA and EAA
So, we can complete stage 1 in 3 ways
This ensures that the three vowels are kept together.

NOTE: We now have the following 6 objects to arrange: P, R, L, L, L and the glued-together-vowels


Stage 2: Arrange the following 6 objects: P, R, L, L, L and the glued-together-vowels
Notice that 3 of our objects are identical L's

----ASIDE------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
-----------------------

In P, R, L, L, L and the glued-together-vowels:
There are 6 objects in total
There are 3 identical L's

So the total number of possible arrangements = 6!/(3!) = 120
So, we can complete stage 2 in 120 ways

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus arrange the letters) in (3)(120) ways (= 360 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GRE. So, be sure to learn it.

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PARALLEL
ALL VOWELS TOGETHER

AAE. PRLLL
(3WORDS)(5words)
(consider 3 together as 1)(5)
total 6words
so
formula=.( total words/repetition in 5) * group1 words/repetition of words in group1

so 6!/3L*3!/2A

6!/3!*3!/2!
=6*5*4*3
360