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M32-05

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M32-05  [#permalink]

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New post 20 Jun 2016, 04:30
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At a birthday party, 10 students are to be seated around a circular table. What is the probability that two of the students, Anna and Bill, do NOT sit next to each other?

A. \(\frac{8}{9}\)
B. \(\frac{7}{9}\)
C. \(\frac{4}{5}\)
D. \(\frac{2}{9}\)
E. \(\frac{1}{9}\)

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Re M32-05  [#permalink]

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New post 20 Jun 2016, 04:30
Official Solution:

At a birthday party, 10 students are to be seated around a circular table. What is the probability that two of the students, Anna and Bill, do NOT sit next to each other?

A. \(\frac{8}{9}\)
B. \(\frac{7}{9}\)
C. \(\frac{4}{5}\)
D. \(\frac{2}{9}\)
E. \(\frac{1}{9}\)


10 students around a circular table can be arranged in \((10-1)!=9!\) ways.

Now, consider Anna and Bill as one unit - {Anna, Bill}. We will have 9 units to arrange: 8 students and {Anna, Bill}. Those 9 units can be arranged around a circular table in \((9-1)!=8!\) ways. Anna and Bill within their unit can be arranged in two ways {Anna, Bill} or {Bill, Anna}. Thus the number of ways to arrange 10 students around a circular table so that two of them, Ana and Bill, sit next to each other is \(8!*2\).

Therefore, the number of ways to arrange 10 students around a circular table so that two of them, Ana and Bill, do NOT sit next to each other is \(9!-8!*2\)

The probability = \(\frac{favorable}{total}=\frac{9!-8!*2}{9!}=1-\frac{2}{9}=\frac{7}{9}\)


Answer: B
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Re: M32-05  [#permalink]

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New post 07 Aug 2016, 23:02
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I solved it in a bit simpler form:

There are 10 spots for 10 people. How can we break the requirement? What's the probability of arranging these two people in a way that they sit next to each other?
Probability of choosing a person is 1, then what's the probability of choosing a person sitting next to him = 1/9 * 2 (since there are two seats right next to him). Then the probability of arranging these two people in a way, that they don't seat to each other is 1 - 2/9 = 7/9
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Re: M32-05  [#permalink]

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New post 07 Aug 2016, 23:13
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manlog wrote:
I solved it in a bit simpler form:

There are 10 spots for 10 people. How can we break the requirement? What's the probability of arranging these two people in a way that they sit next to each other?
Probability of choosing a person is 1, then what's the probability of choosing a person sitting next to him = 1/9 * 2 (since there are two seats right next to him). Then the probability of arranging these two people in a way, that they don't seat to each other is 1 - 2/9 = 7/9


My approach is very similar -

1. Let's say Ann sits at a place.
2. Probability that Bill sits next to Ann in circular arrangement is \(\frac{[Left + Right]}{[Remaining 9 places]}\) = \(\frac{2}{9}\)
3. Probability of both not sitting together is \(1 - \frac{2}{9} = 7/9\)
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Re: M32-05  [#permalink]

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New post 08 Jun 2017, 16:02
This can be solved most easily with a visualization. Create a circle with ten slots. Sit Bill at one randomly. There are now nine slots left. Two are exactly adjacent to Bill. The probability that Ann is placed in one of these seats then is \(\frac{2}{9}\). Thus the probability that Ann isn't next to Bill is \(1-\frac{2}{9}=\frac{7}{9}\)
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Re: M32-05  [#permalink]

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New post 18 Sep 2017, 00:24
why can 10 people be arranged in 9! ways and not in 10! ways?
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Re: M32-05  [#permalink]

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New post 18 Sep 2017, 00:29
ma14292 wrote:
why can 10 people be arranged in 9! ways and not in 10! ways?


The number of arrangements of n distinct objects in a row is given by \(n!\).
The number of arrangements of n distinct objects in a circle is given by \((n-1)!\).

The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have:

\(\frac{n!}{n} = (n-1)!\).

Check other Arrangements in a Row and around a Table questions in our Special Questions Directory.
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New to the Math Forum?
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Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M32-05 &nbs [#permalink] 18 Sep 2017, 00:29
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