M32-05

Question

At a birthday party, 10 students are to be seated around a circular table. Two seating arrangements are considered different only when the positions of the people are different relative to each other. 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. \(\frac{1}{9}\)

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

manhasnoname · Accepted Answer

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{ }{ }  =  \frac{2}{9} 
3. Probability of both not sitting together is  1 - \frac{2}{9} = 7/9

Bunuel · Answer

Official Solution:

At a birthday party, 10 students are to be seated around a circular table. Two seating arrangements are considered different only when the positions of the people are different relative to each other. 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. \(\frac{1}{9}\)

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

There are 10 students around a  circular table , which can be arranged in  (10-1)!=9!  ways.
 
Now, consider Anna and Bill as a single unit - {Anna, Bill}. Then, we have nine units to arrange, namely eight students and {Anna, Bill}. These units can be arranged around the  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 total number of ways to arrange the students such that Anna and Bill sit next to each other is  8!*2 . Therefore, the number of ways to arrange the students so that Anna and Bill do NOT sit next to each other is  9!-8!*2 .
 
The probability of Anna and Bill not sitting next to each other is then  \frac{favorable}{total}=\frac{9!-8!*2}{9!}=1-\frac{2}{9}=\frac{7}{9} .

Answer: B

manlog · Answer

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

spence11 · Answer

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

Attachments

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ma14292 · Answer

why can 10 people be arranged in 9! ways and not in 10! ways?

Bunuel · Answer

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 .

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

sanya511 · Answer

Hey  Bunuel 
Can you explain what the question means by the statement:  Two seating arrangements are considered different only when the positions of the people are different relative to each other 
 I read this and thought we had to do something other than the (n-1) factorial

Bunuel · Answer

That statement just clarifies that rotations of the same arrangement are not counted separately, which is exactly why we use (n - 1)! for circular permutations instead of n!. It doesn’t mean you need a different counting method.

M32-05

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