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

BY-- SRIDHAR S from MANGALORE

9 comments:

  1. C(10,2)ten car taking two at a time
    = 10*9/2
    =45 races
    since If each car must race each of the other cars exactly twice
    45*2=90 races

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  2. 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.

    ReplyDelete
  3. nPr equation.

    nPr= n!/(n-r)!
    =10P2 = 10!/8! = 90

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  4. 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.

    ReplyDelete
  5. B is the correct answer

    ReplyDelete