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13 Feb 2020, 07:48
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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80% (01:09) correct
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Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam and Diane must sit next to each other and Bob and Carol must sit next to each other, how many different seating arrangements are possible on the bench?
A) 3
B) 6
C) 8
D) 12
E) 24
A) 3
B) 6
C) 8
D) 12
E) 24
Updated on: 17 May 2021, 08:36
Originally posted by BrentGMATPrepNow on 13 Feb 2020, 08:01.
Last edited by BrentGMATPrepNow on 17 May 2021, 08:36, edited 1 time in total.
Last edited by BrentGMATPrepNow on 17 May 2021, 08:36, edited 1 time in total.
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Bunuel
Take the task of arranging the people and break it into stages.
Important: Since Adam & Diane must sit next to each other, and since Bob & Carol must sit next to each other, let's "glue" them together to ensure that this happens.
Stage 1: "glue" Adam and Diane together
There are two different ways to glue these two people together: AD or DA.
So, we can complete stage 1 in 2 ways
Stage 2: "glue" Bob & Carol together
There are two different ways to glue these two people together: BC or CB.
So, we can complete stage 2 in 2 ways
At this point we have ensured that the two pairs of people must be next to each other.
We now have 3 "things" to arrange: Ed, the Adam/Diane pair, and the Bob/Carol pair.
Stage 3: Arrange the 3 "things"
We can arrange n unique objects in n! ways.
So we can arrange the 3 " things" in 3! ways (3! = 6)
So, we can complete stage 3 in 6 ways
By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus arrange the 5 people) in (2)(2)(6) ways (= 24 ways)
Answer: E
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.
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13 Feb 2020, 08:15
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First, suppose A and D sit together, in the order AD, and B and C sit together in the order BC. We'd then be making a seating arrangement out of these three things: AD, BC, E. We'd be able to do that in 3! = 6 ways. But for A and D we have two choices -- they could be in the order AD or DA. We also have two choices for the order of B and C. Multiplying our choices, we thus have 6*2*2 = 24 possibilities in total.
