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

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BTGModeratorVI: Legendary Member; Posts: 1223; Joined: Sat Feb 15, 2020 2:23 pmFollowed by: 1 member

by BTGModeratorVI » Sun Jul 26, 2020 6:40 am

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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

Answer: C
Source: Math Revolution

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Source: Beat The GMAT — Problem Solving | Sun Jul 26, 2020 6:40 am

dashingnightmare: Senior | Next Rank: 100 Posts; Posts: 31; Joined: Fri Jul 24, 2020 9:40 pm

Re: Five letters A, P, P, L and E are listed in a row.

by dashingnightmare » Mon Jul 27, 2020 12:55 pm

Total number of arrangement possible without any constraint= $$\frac{5!}{2!}$$ =60
Let 2 P are together, total arrangement =4*3*2*1=24

Total arrangemet-arrangement in which 2 p are together= arrangementswhich have at least one letter between the two Ps
Hence 60-24=36
Hence C.

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Re: Five letters A, P, P, L and E are listed in a row.

by Brent@GMATPrepNow » Tue Jul 28, 2020 5:11 am

BTGModeratorVI wrote: ↑
Sun Jul 26, 2020 6:40 am
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

Answer: C
Source: Math Revolution

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. For more information about the FCP, watch this video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html

MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
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- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html

DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8155; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

Re: Five letters A, P, P, L and E are listed in a row.

by Scott@TargetTestPrep » Fri Jul 31, 2020 7:15 am

BTGModeratorVI wrote: ↑
Sun Jul 26, 2020 6:40 am
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

Answer: C
Source: Math Revolution

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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