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gmatophobia
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


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
­I tried this is the same way but picked 3 and ended up being confused between E and B. solved again with 4 and got it.
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Method 1:
Select 2 cities. For each selection we have 2 flights

NC2*2=n(n-1)

Method 2:
Take n=2 . Say you travel b/w Mumbai and New York.
There are 2 flights (Mumbai to NY & back)

Only option B works

Posted from my mobile device
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Start with two cities. It can have two flights between them. 2*1 = 2
Then , with three cities ; They can have 6 flights among them ; 3*2 = 6

Then with , four cities ; They can 12 flights among them ; 4*3 = 12

Hence using this analogy , we can say that for n cities , we can have n*(n-1) flights.
B is the answer.

KarishmaB , MartyMurray Is my approach correct ?­
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­The simplest way to solve this actually is by using the Permuations formula. Combinatorics creates an extra step by having to multiply your answer by 2 to account for the flights in both directions. 
Perumations already accounts for this since this would be an instance of flight from AB not being the same as flight BA. Combinatorics counts AB the same as BA, ex: strawberry vanilla flavor is the same as vanilla strawberry. 

P(n 2) = n!/(n-2)!= n(n-1)(n-2)!/n-2)! = n(n-1)­
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Start with two cities. It can have two flights between them. 2*1 = 2
Then , with three cities ; They can have 6 flights among them ; 3*2 = 6

Then with , four cities ; They can 12 flights among them ; 4*3 = 12

Hence using this analogy , we can say that for n cities , we can have n*(n-1) flights.
B is the answer.

KarishmaB , MartyMurray Is my approach correct ?­
­Yes its correct. 
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Each city has a code:
for city A and city B, A->B and B->A represents two flights.
So the order of drawing from city list matters.
Simply the logic of An2
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