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04 Mar 2021, 07:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
58% (02:07) correct
42%
(02:19)
wrong
based on 177
sessions
History
Date
Time
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Not Attempted Yet
The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are to be arranged around a circle. What is the probability that all the even numbers are together?
A. 1/9!
B. 1/126
C. 5/126
D. 1/12
E. 2/9
A. 1/9!
B. 1/126
C. 5/126
D. 1/12
E. 2/9
04 Mar 2021, 19:23
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Probability = Favorable Outcomes / Total Outcomes
There are 10 digits to be arranged. These have 5 even numbers.
Favorable Outcomes = Taking all even numbers as 1 unit.
External arrangement = Arranging 5 odd numbers + 1 unit of even numbers = (6 - 1)! = 5!
Internal arrangement = 5! (for all the even numbers)
Total favorable Outcomes = 5! * 5!
Total Arrangements = (10 - 1)! = 9!
The required probability = 5! * 5! / 9! = (5 * 4 * 3 * 2 * 5!) / ( 9 * 8 * 7 * 6 * 5!) = 5 / 126
Option C
Arun Kumar
There are 10 digits to be arranged. These have 5 even numbers.
Favorable Outcomes = Taking all even numbers as 1 unit.
External arrangement = Arranging 5 odd numbers + 1 unit of even numbers = (6 - 1)! = 5!
Internal arrangement = 5! (for all the even numbers)
Total favorable Outcomes = 5! * 5!
Total Arrangements = (10 - 1)! = 9!
The required probability = 5! * 5! / 9! = (5 * 4 * 3 * 2 * 5!) / ( 9 * 8 * 7 * 6 * 5!) = 5 / 126
Option C
Arun Kumar
13 Feb 2024, 08:10
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CrackverbalGMAT
Why are the total arrangements (10 -1)! ?
I took as total arrangements just 10! ...
