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Monday, June 11, 2007
GRE QUESTION ON DEMAND
In a series of races, 10 toy cars are raced, 2 cars at a time. If each car must race each of the other cars exactly twice, how many races must be held? (A) 40 (B) 90 (C) 100 (D) 180 (E) 200
it is D,180. we can select 2 out of 10 cars in 10c2 = 45 ways. now,each car must race each other car,hence order becomes important,so number of races 2 b held =45*2=90. but each car must race each other twice,so twice again gives 180 as da answer. guys n gals,a better explanation wud always b welcome.
I think Miss Kiran is wrong.What order are we talking about.It is a race b/w two cars and not which one among the two can be first or second.The result is immaterial. If car 1 races aginst car 2..It is the same as car 2 racing against car 1 irrespective of the fact that which car wins the race.
b
ReplyDeleteC(10,2)ten car taking two at a time
ReplyDelete= 10*9/2
=45 races
since If each car must race each of the other cars exactly twice
45*2=90 races
B
ReplyDeleteit is D,180.
ReplyDeletewe can select 2 out of 10 cars in 10c2 = 45 ways.
now,each car must race each other car,hence order becomes important,so number of races 2 b held =45*2=90.
but each car must race each other twice,so twice again gives 180 as da answer.
guys n gals,a better explanation wud always b welcome.
nPr equation.
ReplyDeletenPr= n!/(n-r)!
=10P2 = 10!/8! = 90
I think Miss Kiran is wrong.What order are we talking about.It is a race b/w two cars and not which one among the two can be first or second.The result is immaterial.
ReplyDeleteIf car 1 races aginst car 2..It is the same as car 2 racing against car 1 irrespective of the fact that which car wins the race.
b
ReplyDeleteB is the correct answer
ReplyDeleteb is the correct
ReplyDelete