Five letters A, P, P, L and E are listed in a row. How many arrangemen

Here's an approach that doesn't require us to subtract the bad arrangements.

Take the task of arranging the 5 letters and break it into stages.

Stage 1: Arrange the letters A, L, E in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 3 letters in 3! ways (= 6 ways)
So, we can complete stage 1 in 6 ways

IMPORTANT: For each arrangement of 3 letters (above), there are 4 places where the two P's can be placed.
For example, in the arrangement AEL, we can add spaces as follows _A_E_L_
So, if we place each P in one of the available spaces, we can ENSURE that the two P's are never together.

Stage 2: Select two available spaces and place an P in each space.
Since the order in which we select the two spaces does not matter, we can use combinations.
We can select 2 spaces from 4 spaces in 4C2 ways (= 6 ways)
So we can complete stage 2 in 6 ways.

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus arrange all 5 letters) in (6)(6) ways (= 36 ways)

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So be sure to learn this technique.

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When we encounter “at least” in counting questions or probability questions, we should consider complementary counting.
The total number of arrangements of the 5 letters is 5!/2! (5! Counts each arrangement of the two Ps 2! times).
The number of arrangements with no letter between the two Ps is 4!.
Thus, the number of arrangements in which at least one letter lies between the two Ps is 5!/2! – 4! = 60 – 24 = 36.

Therefore, C is the answer.
Answer: C

Solution:

We use the indistinguishable permutations formula (because of the two occurrences of letter P) to calculate the total number of arrangements without any restrictions: 5!/2! = 120/2 = 60. The total number of arrangements where the two P’s must be together (i.e., no letters can be between them) is 4! = 24. Since all the other arrangements will have at least one letter between the two P’s, the number of such arrangements is 60 - 24 = 36.

Answer: C

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