Medals are to be awarded to three teams in a 10-team : GMAT Problem Solving (PS)
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# Medals are to be awarded to three teams in a 10-team

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Medals are to be awarded to three teams in a 10-team [#permalink]

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02 Oct 2010, 02:54
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Medals are to be awarded to three teams in a 10-team competition. If one medal is gold, one medal is silver, and one medal is bronze, how many different ways are there to award the three medals to teams in the competition?

A. 10!/7!
B. 10!/(3!7!)
C. 10!/3!
D. 7!/3!
E. 7!/94!3!)
[Reveal] Spoiler: OA
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02 Oct 2010, 03:46
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anilnandyala wrote:
Medals are to be awarded to three teams in a 10-team competition. If one medal is gold, one medal is silver, and one medal is bronze, how many different ways are there to award the three medals to teams in the competition?

a) 10!/7!
b)10!/3! 7!
c)10!/3!
d)7!/3!
e)7!/4! 3!

Choosing 3 teams out of 10 when order of the teams matters - $$P^3_{10}=\frac{10!}{7!}$$;

Or: choosing which 3 teams out of 10 will get the medals - $$C^3_{10}$$ and arranging them - $$3!$$, so total - $$C^3_{10}*3!=\frac{10!}{7!}$$;

Or:
1-2-3-4-5-6-7-8-9-10 (teams);
G-S-B-N-N-N-N-N-N-N (GSB - medals, N - no medal);

Permutation of 10 letters out of which 7 N's are identical is $$\frac{10!}{7!}$$ (so you'll get $$\frac{10!}{7!}$$ different ways of assigning the medals to the teams).

Hope it's clear.
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Re: medals are to be aworded [#permalink]

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02 Oct 2010, 07:15
anilnandyala wrote:
Medals are to be awarded to three teams in a 10-team competition. If one medal is gold, one medal is silver, and one medal is bronze, how many different ways are there to award the three medals to teams in the competition?

a) 10!/7!
b)10!/3! 7!
c)10!/3!
d)7!/3!
e)7!/4! 3!

Step 1 : Choose 3 teams out of 10 = C(10,3)
Step 2 : Distribute the 3 medals between these = 3!

Answer : C(10,2) * 3! = 10!/7! or (A)
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03 Oct 2010, 02:52
its choosing 3 teams from 10 teams when the order matters
so its 10P3.
the ans is A
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Re: medals are to be aworded [#permalink]

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03 Oct 2010, 08:44
Hi Bunuel,

Pls tell me if there is anything wrong with my approach...

10C1*9C1*8C1 which comes out to be A
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Re: medals are to be aworded [#permalink]

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20 Oct 2010, 04:34
I thought the ten choose three formula was (n choose k)= n!/(k!(n-k)!)

Am I over thinking this...
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Re: medals are to be aworded [#permalink]

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23 Feb 2011, 13:45
georgea wrote:
I thought the ten choose three formula was (n choose k)= n!/(k!(n-k)!)

Am I over thinking this...

Your formula only works when the order does not matter.
When the order matters, as it does in this case, the formula is: (n Permut k)= n!/(n-k)!
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Re: medals are to be aworded [#permalink]

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23 Feb 2011, 20:51
Agreed, A. I used slot method as order matters
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Re: Medals are to be awarded to three teams in a 10-team [#permalink]

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19 Aug 2014, 22:44
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Re: Medals are to be awarded to three teams in a 10-team   [#permalink] 19 Aug 2014, 22:44
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