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18 Jul 2022, 08:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
54% (02:04) correct
46%
(01:55)
wrong
based on 200
sessions
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Not Attempted Yet
11 players of the Portugal National Football Team are standing in a row in random order. If no two players of the Portugal National Football Team have the same name, what is the probability that by switching exactly two players with each other, the players will be standing in alphabetical order of their names ?
A. \(\frac{1}{11!}\)
B. \(\frac{2}{11!}\)
C. \(\frac{5}{11!}\)
D. \(\frac{55}{11!}\)
E. \(\frac{110}{11!}\)
A. \(\frac{1}{11!}\)
B. \(\frac{2}{11!}\)
C. \(\frac{5}{11!}\)
D. \(\frac{55}{11!}\)
E. \(\frac{110}{11!}\)
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18 Jul 2022, 08:41
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Bunuel
On questions like this, I like to give myself a (much) easier example to make sure I've got the setup right. Let's try it with four players: A, B, C, and D.
How many total ways are there to have them lined up? 4! = 24
How many ways are there to line them up so that we can do the one swap and get to alphabetical?
We could trade A with any of the others: BACD, CBAD, DBCA. 3 ways.
We could trade B with any of the others aside from A because we've already accounted for the AB swap: ACBD, ADCB. 2 ways.
We could trade C with any of the others aside from A or B because we've already accounted for the AC and BC swaps: ABDC. 1 way.
That's 6 ways.
So, our setup is [(n-1)+(n-2)+(n-3)...]/n!
\(\frac{10+9+8+7+6+5+4+3+2+1}{11!}\)
\(\frac{55}{11!}\)
Answer choice D.
General Discussion
18 Jul 2022, 08:42
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Bunuel
The probability that all the 11 players are standing alphabetically = 1/11! (only 1 arrangement out of 11!).
We have to find the probability that by switching exactly two players with each other, the players will be standing in alphabetical order of their names.
It means only 2 players need to be selected and arranged in a defined manner. It will be done by 11C2 = 11*10*9!/9! * 2 = 55
Probability = 55/11!
Hence Option D is the answer.
