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
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9 comments:
b
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
B
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.
nPr equation.
nPr= 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.
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
B is the correct answer
b is the correct
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