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The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are to be arranged around

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Bunuel
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The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are to be arranged around [#permalink] The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are to be arranged around   04 Mar 2021, 06:15
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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
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chetan2u
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Re: The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are to be arranged around [#permalink] The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are to be arranged around   04 Mar 2021, 20:36
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Bunuel wrote:
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



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 = \(1-\frac{2}{9}=\frac{7}{9}\)

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Re: The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are to be arranged around [#permalink] The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are to be arranged around   12 Mar 2021, 10:43
Bunuel wrote:
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


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.
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Re: The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are to be arranged around [#permalink] The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are to be arranged around   17 Jun 2021, 06:06
Total number of arrangements of ten digits around a circle is (10-1)!=9!

As digits 8 and 9 cannot be next to each other
1. Arrange nine digits (0,1,2,3,4,5,6,7,8) except digit 9 around a circle in (9-1)!=8! number of ways
2. There are nine spaces between nine digits arranged around a circle but only seven are available for digit 9 so that it is not next to digit 8
Total number of possible arrangements = 8!×7

8!×7/9!=7/9
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Re: The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are to be arranged around [#permalink] The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are to be arranged around   17 Jun 2021, 07:07
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Imagine putting the '8' somewhere on the circle first. There now remain 9 slots around the circle, and 7 of those slots are not next to the '8', so if we put the '9' somewhere at random, the probability is 7/9 it is not next to the '8'.
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Re: The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are to be arranged around [#permalink]
17 Jun 2021, 07:07
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