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07 Apr 2026, 18:21
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Difficulty:
55%
(hard)
Question Stats:
56% (01:37) correct
44%
(01:41)
wrong
based on 34
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Five different surnames are listed in a random order. What is the probability that the names can be made to be in alphabetical order if exactly two names are swapped?
A) \(\frac{1}{15}\)
B) \(\frac{1}{12}\)
C) \(\frac{1}{10}\)
D) \(\frac{1}{8}\)
E) \(\frac{1}{6}\)
A) \(\frac{1}{15}\)
B) \(\frac{1}{12}\)
C) \(\frac{1}{10}\)
D) \(\frac{1}{8}\)
E) \(\frac{1}{6}\)
08 Apr 2026, 09:47
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Five different surnames are listed in a random order. What is the probability that the names can be made to be in alphabetical order if exactly two names are swapped?
Find total number of arrangements of the names:
5! = 120
Find number of ways they can be in alphabetical order:
1
Find number of ways exactly two can be swapped from alphabetical order to not alphabetical order:
5c2 = (5 × 4)/2! = 10
The probability that exactly two are swapped from alphabetical order to not alphabetical order is the same as the probability that two can be swapped to alphabetical order:
(1 × 10)/120 = 1/12
A) \(\frac{1}{15}\)
B) \(\frac{1}{12}\)
C) \(\frac{1}{10}\)
D) \(\frac{1}{8}\)
E) \(\frac{1}{6}\)
Correct answer: B
Find total number of arrangements of the names:
5! = 120
Find number of ways they can be in alphabetical order:
1
Find number of ways exactly two can be swapped from alphabetical order to not alphabetical order:
5c2 = (5 × 4)/2! = 10
The probability that exactly two are swapped from alphabetical order to not alphabetical order is the same as the probability that two can be swapped to alphabetical order:
(1 × 10)/120 = 1/12
A) \(\frac{1}{15}\)
B) \(\frac{1}{12}\)
C) \(\frac{1}{10}\)
D) \(\frac{1}{8}\)
E) \(\frac{1}{6}\)
Correct answer: B
