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Probability glitch!

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Probability glitch!

by ajas » Sun Apr 18, 2010 9:57 am
Can someone please solve and explain the following question to me?

At a dinner party 5 people are to be seated around a circular table. Two seating arragements are considered different only when the positions of the people are different relative to each other.what is the total number of different possible seating arrangements for the group?

A. 5
B. 10
C. 24
D. 32
E. 120

Thanks!
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by eaakbari » Sun Apr 18, 2010 10:25 am
Circular permutation = (n-1)!
= 4!
= 24
Answer C
Last edited by eaakbari on Sun Apr 18, 2010 1:07 pm, edited 1 time in total.
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by chrsrook » Sun Apr 18, 2010 12:48 pm
circulation permutation- n!/n

5!/5=24

C

Thanks
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by ajas » Sun Apr 18, 2010 2:35 pm
thanks both of you!
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by tanviet » Sun Apr 18, 2010 11:22 pm
Please, help

for circular permutation. How do we have

(N-1)!
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by liferocks » Sun Apr 18, 2010 11:24 pm
duongthang wrote:Please, help

for circular permutation. How do we have

(N-1)!
good explanation is provided here.Please check
https://www.beatthegmat.com/circular-per ... tml#241159
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by ajith » Sun Apr 18, 2010 11:27 pm
duongthang wrote:Please, help

for circular permutation. How do we have

(N-1)!

Say we want to arrange A B C in normal permutation

ABC ACB BAC BCA CAB CBA - There are 6 ways to arrange A B C

or n! ways (n being 3)

Now in a circular permutation ABC, BCA and CAB are the same
Also ACB, CBA and BAC are the same (because it is circular and hence the start and the end doesnt matter the order matters)

There are only 2 permuations

or (n-1)! (n being 2)

That's one way to explain why (n-1)!
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by eaakbari » Mon Apr 19, 2010 12:34 am
When you arrange n objects linearly you have n!
When you arrange them in a circular fashion, that time abc is the same thing as cba when n = 3.
In general you will have n number of redundancies because of circular perm.

Hence for circular = n!/n = (n-1)!
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