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27 Jan 2024, 10:03
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Difficulty:
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Question Stats:
69% (01:47) correct
31%
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An airline serving n cities has 2 nonstop flights per day, one in each direction, between each pair of these cities. How many nonstop flights does the airline have among these n cities each day?
A. 2n(n-1)
B. n(n-1)
C. 2n^2
D. n^2
E. 2n
2024-01-27_20-58-23.png [ 37.25 KiB | Viewed 6720 times ]
A. 2n(n-1)
B. n(n-1)
C. 2n^2
D. n^2
E. 2n
Attachment:
2024-01-27_20-58-23.png [ 37.25 KiB | Viewed 6720 times ]
27 Jan 2024, 10:11
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guddo
An airline serving n cities has 2 nonstop flights per day, one in each direction, between each pair of these cities. How many nonstop flights does the airline have among these n cities each day?
A. 2n(n-1)
B. n(n-1)
C. 2n^2
D. n^2
E. 2n
A. 2n(n-1)
B. n(n-1)
C. 2n^2
D. n^2
E. 2n
Attachment:
2024-01-27_20-58-23.png
From n cities, we can form \(C^2_n=\frac{n!}{2!(n-2)!}=\frac{(n-2)!(n-1)n}{2!(n-2)!}=\frac{(n-1)n}{2}\) pairs. Since there are 2 flights between each pair of cities, the total number of flights among n cities is \(2*\frac{(n-1)n}{2}=(n-1)n\).
Answer: B.
General Discussion
27 Jan 2024, 11:24
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guddo
An airline serving n cities has 2 nonstop flights per day, one in each direction, between each pair of these cities. How many nonstop flights does the airline have among these n cities each day?
A. 2n(n-1)
B. n(n-1)
C. 2n^2
D. n^2
E. 2n
A. 2n(n-1)
B. n(n-1)
C. 2n^2
D. n^2
E. 2n
Attachment:
2024-01-27_20-58-23.png
One way to solve this question is by using the concept of Permutations and Combinations as Bunuel has depicted.
Another way is to assume a value of n and then eliminate the options.
Assume n = 2
The number of flights nonstop flights the airline has among these 2 cities ⇒ is 2 (one flight from City 1 to City 2 and another from City 2 to City 1
Answer choice elimination
A. 2n(n-1) ⇒ 2 * 2 (1) = 4
B. n(n-1) ⇒ 2 * 1 = 2
C. 2n^2 ⇒ 2*2^2 = 8
D. n^2 ⇒ 2^2 = 4
E. 2n ⇒ 2 * 2 = 4
Option B
