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Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam an

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Bunuel
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Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam an [#permalink] New post   26 Apr 2017, 08:34
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Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam and Diane cannot sit next to each other, how many different seating arrangements are possible on the bench?

A. 5
B. 10
C. 48
D. 72
E. 120
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D
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Re: Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam an [#permalink] New post   26 Apr 2017, 10:54
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Total number of ways all 5 can sit = 5! = 120
Total number of ways Adam and Diane can sit together = 4!*2 = 24*2 = 48
Total number of ways Adam and Diane cannot sit together = 120 - 48 = 72

Answer = D. 72
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Re: Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam an [#permalink] New post   27 May 2017, 01:39
the answer should be D
adam can sit in 4 ways likewise diane could sit in 3 ways rest can be sit in 3! ways so toatal will be 4*3*3!=72
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Re: Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam an [#permalink] New post   14 Aug 2021, 21:47
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Total people: A,B,C,D, and E = 5

The number of ways to arrange 5 people = 5! = 120

If A and D are together: B, C, E, (A, D):

Total people: 4 and can be arranged in 4! ways * (A,D) arranged in 2! ways = 24 * 2 = 48

Total ways in which A and D are not together: 120 - 48 = 72

Answer D
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Re: Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam an [#permalink] New post   14 Aug 2021, 23:04
Bunuel wrote:
Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam and Diane cannot sit next to each other, how many different seating arrangements are possible on the bench?

A. 5
B. 10
C. 48
D. 72
E. 120


let us first arrange all of them in 5! ways

then eleminating the possibilities of Adam and Diane together =4!*2!

Therefore the total ways of arrangement = 120-48
=>72

Therefore IMO D
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Re: Adam, Bob, Carol, Diane, and Ed are all sitting on a bench. If Adam an [#permalink]
14 Aug 2021, 23:04
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