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Circular Permutation?? |
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| Nov24-10, 05:45 PM | #1 |
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Circular Permutation??
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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| Nov25-10, 08:01 AM | #2 |
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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!
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| Nov25-10, 08:17 AM | #3 |
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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) |
| Nov25-10, 10:54 AM | #4 |
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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 |
| Nov25-10, 11:25 AM | #5 |
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My answer:
[tex]\frac{( 21 +15) \cdot 5! \cdot 7!}{12} = 1.814.400 [/tex] (this equals you answer) |
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| combination, permutation, probabality, set theory |
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