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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Mar 2007 14:22:07 IST
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arey ( I THINK) akela ek banda to speech nahi dega na baaki 7 bhi to de sakte hain
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I always like to walk in rain as no one can see me crying there :(
frnds are like diamonds , if u hit them , they don't break but they slip frm ur hands
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Mar 2007 14:37:42 IST
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but in the ques it is written specific person
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Mar 2007 16:53:42 IST
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the ans is B...bt can any1 plz explain in detail
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nupur.. |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Mar 2007 04:20:36 IST
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answer is simply (b).
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Mar 2007 09:26:06 IST
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Hi The ans can be b) only if the speaker's are not specified because then we will get 8C2 * 6! =20160 but if the speaker's are specified(i.e selections of the two speakers are not to be made) as given in the question, then the ans would be different
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Mar 2007 12:44:47 IST
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the ans is definately b
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Mar 2007 13:25:42 IST
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how?? plz explain.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Mar 2007 13:44:31 IST
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Its (b) without any restriction the 8 people can be arranged themselves in 8! ways. But the number of ways in which A speaks up before B and the number of ways in which B speaks before A together make up 8!. (Number of ways in which A speaking before B is equal to the number of ways B speaking before A) so the required number of ways= 8!/2 = 20160 plz rate me...
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Mar 2007 13:51:26 IST
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thanx a lot!!!! but could u plz tell where i went wrong.???(my method is on page 1)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Mar 2007 14:36:28 IST
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as there are 8 persons and as one person has to speak before aspecified person his probability is half, i e 8!/2 = 20160
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Mar 2007 15:21:32 IST
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hi abijeet the question says that both A and B are specific. it means A have to come first and after that it should be B and no one else. if we go for 7! ways after placing A as you did, it will also include B being some where else. hope u getting me.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Mar 2007 15:50:57 IST
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ok thanks i didnt understand the ques but if B has to come immediately after A,y cant v do it like this let A & B be grouped together now we hav 7 person which can b arranged in 7! ways. but then the ans comes 5040.what's the mistake?????
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 Mar 2007 13:19:59 IST
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suppose speaker A had to speak before B.The remaining 6 speakers can take any of the 6positions out of total 8.they can be arranged in 6! ways. when A occupies first position,B can occupy any one of the other 7 positions. Total no.of ways in which A is in post1 and B in other seven is - 7^6!. when Ais in post2,B has 6 posts.remaining are arranged in 6! ways Total no of ways-6^6! when A is in3 B has 5 posts. total ways-5^6! continuing this way until A is in pos7 and B is in pos8 total ways=6!(7+6+5+4+3+2+1)=20160 ansb
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 Mar 2007 14:43:22 IST
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hi answer should be:- "d"
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2007 14:14:16 IST
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hi nikhil how is it d
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