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Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam an
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13 Feb 2020, 06:48
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55% (hard)
Question Stats:
73% (01:28) correct 27% (00:52) wrong based on 26 sessions
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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?
Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam an
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13 Feb 2020, 07:01
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Bunuel wrote:
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
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.
Re: Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam an
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13 Feb 2020, 07:15
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.
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Re: Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam an
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14 Feb 2020, 07:00
Bunuel wrote:
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
for A& B ( X) ; 2 ways and for C&D (Y) ; 2 ways now XYE can be arranged in 3! ways so total ways possible 2*2*6 ; 24 IMO E
gmatclubot
Re: Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam an
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14 Feb 2020, 07:00