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

Question

Five letters A, P, P, L and E are listed in a row. How many arrangements have at least one letter between the two Ps?

A. 24 
B. 30 
C. 36 
D. 42 
E. 48

GMATGuruNY · Accepted Answer

Good arrangements = total arrangements - bad arrangements.

Total arrangements: 
Number of ways to arrange 5 elements = 5!. 
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED. 
The reason: 
When the identical elements swap positions, the arrangement doesn't change. 
Here, we must divide by 2! to account for the two identical P's: 
5!/2! = 60.

Bad arrangements: 
In a bad arrangement, the two P's are in adjacent slots. 
Let   represent the 2 adjacent P's. 
Number of ways to arrange the 4 elements  , D, G and T = 4! = 24.

Good arrangements: 
Total arrangements - bad arrangements = 60-24 = 36.

C

GMATinsight · Answer

Total Arrangements of all letters = 5!/2! = 60
Total Arrangements of all letters with two Ps together = 4! = 24

Total arrangements with two Ps together (i.e. atleast one letter between two Ps) = 60-24 = 36

Answer: Option C

BrentGMATPrepNow · Answer

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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MathRevolution · Answer

=>

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

ScottTargetTestPrep · Answer

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

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