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Five stand-by passengers are waiting for 3 open seats on an airplane f
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16 Jan 2023, 02:29
16 Jan 2023, 02:29
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Question Stats:
62% (00:45) correct
37% (00:35) wrong based on 82 sessions
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Five stand-by passengers are waiting for 3 open seats on an airplane flight. In how many different ways can 3 passengers be arranged in these seats?
(A) 10
(B) 15
(C) 20
(D) 60
(E) 125
(A) 10
(B) 15
(C) 20
(D) 60
(E) 125
Re: Five stand-by passengers are waiting for 3 open seats on an airplane f
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18 Jan 2023, 05:00
18 Jan 2023, 05:00
Expert Reply
OE
You can solve this problem by using the Anagram method. Five people (call them A through E) will be assigned to seats (designated as 7, 2, and 3), with 2 people left without a seat (designated as N)
screenshot.1180.jpg [ 13 KiB | Viewed 3304 times ]
The total number of possibilities is the number of arrangements of the “word” \(123 N N\), which yields the same formula as above: 5 items, so 5!, divided by a factorial for all “repeats.” The N is listed twice, so the correct expression is 5!/2!
You can solve this problem by using the Anagram method. Five people (call them A through E) will be assigned to seats (designated as 7, 2, and 3), with 2 people left without a seat (designated as N)
Attachment:
screenshot.1180.jpg [ 13 KiB | Viewed 3304 times ]
The total number of possibilities is the number of arrangements of the “word” \(123 N N\), which yields the same formula as above: 5 items, so 5!, divided by a factorial for all “repeats.” The N is listed twice, so the correct expression is 5!/2!
Re: Five stand-by passengers are waiting for 3 open seats on an airplane f
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01 Oct 2025, 22:26
01 Oct 2025, 22:26
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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