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Going for gold
gmat_aug
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How many ways are there to award a gold, silver and bronze medal to 10 contending teams?
A) 10 × 9 × 8
B) 10! / (3! 7!)
C) 10! / 3!
D) 360
E) 300

I think A and C are same. NOt sure if I am missing something obvious. Can we get such question in GMAT when 2 values are represented differently?

Any early reply will help as taking GMAT this weekend. Thanks
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they are not same.
10.9.8 is equal to 10!/(10-3)!
Re: Going for gold
Ron Purewal
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Joined: 08 Oct 2007
Posts: 2366

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gmat_aug wrote:
How many ways are there to award a gold, silver and bronze medal to 10 contending teams?
A) 10 × 9 × 8
B) 10! / (3! 7!)
C) 10! / 3!
D) 360
E) 300

I think A and C are same. NOt sure if I am missing something obvious. Can we get such question in GMAT when 2 values are represented differently?

Any early reply will help as taking GMAT this weekend. Thanks


if you expand out (c), you'll notice that it differs from (a):
(c) = (10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (3 x 2 x 1) = 10 x 9 x 8 x 7 x 6 x 5 x 4
that's not the same as (a).

btw, it's not unheard of to have two answers that wind up being equivalent. if that's the case, then both are wrong - because, if one of them were right, then the other one would be right too, and you can't have 2 right answers.
cramya
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I think its A) 10*9*8 which 10p3
Ron Purewal
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Joined: 08 Oct 2007
Posts: 2366

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cramya wrote:
I think its A) 10*9*8 which 10p3


i guess i forgot to mention this, heh heh.
yes, the correct answer is (a). because there are three distinct medals being awarded, you can simply multiply the numbers of consecutive possibilities, without having to divide by any "redundancy factor" factorials.
Going for gold
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