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Going for gold

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gmat_aug
 Post subject: Going for gold
 Post Posted: Thu Aug 07, 2008 1:35 pm 
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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Guest
 Post subject:
 Post Posted: Sat Aug 09, 2008 1:33 pm 
they are not same.
10.9.8 is equal to 10!/(10-3)!
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RonPurewal
 Post subject: Re: Going for gold
 Post Posted: Thu Sep 04, 2008 7:46 am 
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ManhattanGMAT Staff


Posts: 7146
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.
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cramya
 Post subject:
 Post Posted: Fri Sep 05, 2008 7:38 pm 
I think its A) 10*9*8 which 10p3
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RonPurewal
 Post subject:
 Post Posted: Thu Oct 09, 2008 6:26 am 
Offline
ManhattanGMAT Staff


Posts: 7146
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.
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