Circular Permutation??


by jxta
Tags: combination, permutation, probabality, set theory
jxta
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#1
Nov24-10, 05:45 PM
P: 2
if there are 7 boys and 5 girls, how many circular arrangements are possible if the ladies do not sit adjacent to each other.??
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tiny-tim
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#2
Nov25-10, 08:01 AM
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hi jxta! welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help!
jxta
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#3
Nov25-10, 08:17 AM
P: 2
Quote Quote by tiny-tim View Post
hi jxta! welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help!
i think :-

boys ways;-(7-1)=6!
now there are 5 girls and 7 seats(in b/w boys) so there are P(7,5) number of ways, the girls can sit.
p(7,5)=7!/(7-5)!

i.e, total no. of ways= 6!*p(7,5)
= 6!*7!/(7-5)!
= 1814400 (but this ans is wrong).

ans = 252 (in my book)

Outlined
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#4
Nov25-10, 10:54 AM
P: 124

Circular Permutation??


you have to divide by 12 (and not 2 * 12 = 24 as you can not mirror) at some step, as it is a circular placement.

252 is definitely wrong, look at the following (non-circular) configuration:

B g B g B g B g B g B B

This gives us 5! * 7! = 604.800 possibilities. Divide by 12 gives 50.400 possibilities. So the answer must be greater than (or equal to) 50.400
Outlined
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#5
Nov25-10, 11:25 AM
P: 124
My answer:

[tex]\frac{( 21 +15) \cdot 5! \cdot 7!}{12} = 1.814.400 [/tex]

(this equals you answer)


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