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![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Mar 2008 21:47:08 IST
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Q-Out of the m persons sitting around a round table , the persons A,B,C are selected at random.THEN THE CHANCE THAT NO TWO OF THEM ARE SITTING TOGETHER IS?
ans-(m-4)(m-5)/(m-1)(m-2)
i m getting (m+1)(m-4)/(m-1)(m-2)
using 1- 3!(m-3)!/(m-1)!
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25 Mar 2008 22:52:07 IST
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![\mbox{Let the first person be selected} \\ \\ \mbox{Now, the next 2 persons can be selected out of the remaining m-3 in} \\ \\ (m-3)C_2 \ \mbox{ways.But there will be m-4 ways in which they get selected together} \\ \\ \mbox{So we subtract them} \\ \\ (m-3)C_2 - (m-4) \\ \\ \mbox{But this can happen to any of the m persons.So we multiply by m} \\ \\ m[(m-3)C_2 - (m-4)] \\ \\ \mbox{But each combination we are counting thrice.So divide by 3} \\ \\ \frac{m[(m-3)C_2 - (m-4)]}{3} \\ \\ \mbox{This is the numerator} \\ \\ \mbox{Since order is not important we get} \ mC_3 \\ \\ \mbox{Thus dividing we get} \\ \\ \frac{(m-4)(m-5)}{(m-1)(m-2)}](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/5/3/7/5376f5f53944ee670ed9dfbd98351d39896a80f7.gif)
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this reply: 20 points
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26 Mar 2008 00:54:44 IST
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i didn't understand ur solution...
but wats wrong in my answer?
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26 Mar 2008 12:11:30 IST
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got it!!
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