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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
Expert Reply
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Question Stats:
65% (00:37) correct
34% (00:26) wrong based on 29 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 900 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 900 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!
gmatclubot
Re: Five stand-by passengers are waiting for 3 open seats on an airplane f [#permalink]
18 Jan 2023, 05:00
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