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04 Mar 2021, 06:15
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
70% (01:48) correct
30%
(01:48)
wrong
based on 216
sessions
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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 the number 8,9 are separated ?
A. 1/9!
B. 1/126
C. 5/126
D. 1/12
E. 7/9
A. 1/9!
B. 1/126
C. 5/126
D. 1/12
E. 7/9
04 Mar 2021, 20:36
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Bunuel
Total ways 10 digits can be arranged around a circular table = (10-1)! or 9!
Let us find ways 8,9 are together, and subtract from total.
Take these 2 digits together, so number of digits =9.
Ways to arrange these 9 digits around a circular table =8!
However, 8,9 can be arranged in 2! ways => 8!*2
P=\(\frac{8!*2}{9!}=\frac{2}{9}\)
So P of 8,9 not being together = \(1-\frac{2}{9}=\frac{7}{9}\)
E
General Discussion
12 Mar 2021, 10:43
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Bunuel
Consider 8 and 9 as 1 unit
So now we have a total of 9 units.
0, 1, 2, 3, 4, 5, 6, 7, {8,9}
We can arrange them in a circle in (n-1)! ways i.e. 8! ways further 8 and 9 can be interchanged hence 2! ways.
Hence total ways when 8 and 9 are together is 8!2!
Number of total ways of arranging in a circle when 8 and 9 are NOT considered as ONE unit = 9! ways
So total ways - total ways {8,9} are together = Total ways 8 and 9 are separated.
probabiity:
\(=\frac{9! -8!2!}{9!}\)
\(=\frac{9*8!-8!2!}{9*8!}\)
\(=\frac{7}{9}\)
Ans-E
Hope it's clear.
