
#1
Nov2410, 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.??




#2
Nov2510, 08:01 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

hi jxta! welcome to PF!
Show us what you've tried, and where you're stuck, and then we'll know how to help! 



#3
Nov2510, 08:17 AM

P: 2

boys ways;(71)=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!/(75)! i.e, total no. of ways= 6!*p(7,5) = 6!*7!/(75)! = 1814400 (but this ans is wrong). ans = 252 (in my book) 



#4
Nov2510, 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 (noncircular) 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 



#5
Nov2510, 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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