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# M32-05

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Math Expert
Joined: 02 Sep 2009
Posts: 42352

Kudos [?]: 133301 [0], given: 12445

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20 Jun 2016, 04:30
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Difficulty:

65% (hard)

Question Stats:

41% (00:48) correct 59% (01:21) wrong based on 32 sessions

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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. $$\frac{1}{9}$$
[Reveal] Spoiler: OA

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Kudos [?]: 133301 [0], given: 12445

Math Expert
Joined: 02 Sep 2009
Posts: 42352

Kudos [?]: 133301 [0], given: 12445

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20 Jun 2016, 04:30
Expert's post
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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 = $$\frac{favorable}{total}=\frac{9!-8!*2}{9!}=1-\frac{2}{9}=\frac{7}{9}$$

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Manager
Joined: 08 Jan 2015
Posts: 86

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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

Kudos [?]: 10 [2], given: 53

Manager
Joined: 21 Apr 2016
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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 $$1 - \frac{2}{9} = 7/9$$

Kudos [?]: 17 [1], given: 79

Intern
Joined: 06 Apr 2017
Posts: 27

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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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Intern
Joined: 05 Sep 2017
Posts: 4

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18 Sep 2017, 00:24
why can 10 people be arranged in 9! ways and not in 10! ways?

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Math Expert
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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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Re: M32-05   [#permalink] 18 Sep 2017, 00:29
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# M32-05

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