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by pemdas » Wed Feb 15, 2012 5:03 pm
i finally revised everything in my made up question and see now that the discrepancy took place with Demorgan's and you are perfectly correct opening up avenues for updating one of the answer choices with your answer. Thank you Mike.

Indeed it's or/and boolean relationship which makes sense here and the double counting which you finely explain must be avoided between two subsets, could be 4!*4 too - yet you applied arrangements and showed symmetry applications there. Just perfect.
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by pappueshwar » Tue Feb 21, 2012 7:25 am
dhiren8182 wrote:The way i solved it!!!

Total no of ways to arrange 6 people =6!
Condition:
Mike and Stuart are not sitting together
Anurag and Ron are not sitting together .
Consider arranging Mitch and Rich in 2! ways+considering M & S as 1 group+considering A & S as 1 group .
Therefore total no of ways is 4!
M & S and A & R can be arranged in 2 ! ways each .
So the arrangement becomes 4*4!.
No of ways they are not sitting together=6!-4*4!=4!(6*5-4)
=26*24=624
Ans is D
hi dhiren,

úr solution is easy to understand but tell me how did u land up at 4 * 4! ? from where did u get that 4?
2 ! is 2 right?