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Medals are to be awarded to three teams in a 10-team

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anilnandyala
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Medals are to be awarded to three teams in a 10-team [#permalink] New post   02 Oct 2010, 03:54
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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!)
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A
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Bunuel
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Re: medals are to be aworded [#permalink] New post   02 Oct 2010, 04:46
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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!

pls provide answer with explanation


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).

Answer: A.

Hope it's clear.
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Re: medals are to be aworded [#permalink] New post   02 Oct 2010, 08:15
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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!

pls provide answer with explanation


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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Re: medals are to be aworded [#permalink] New post   03 Oct 2010, 03:52
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] New post   03 Oct 2010, 09:44
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] New post   20 Oct 2010, 05:34
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] New post   23 Feb 2011, 14:45
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] New post   23 Feb 2011, 21:51
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] New post   08 Feb 2020, 13:19
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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!/94!3!)


Since the order matters (i.e., who gets gold, who gets silver and who gets bronze matters), the number of ways that 3 teams can be chosen from 10 teams is:

10P3 = 10 x 9 x 8 = (10 x 9 x 8 x 7!)/7! = 10!/7!

Answer: A
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Re: Medals are to be awarded to three teams in a 10-team [#permalink] New post   09 Feb 2020, 06:15
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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!/94!3!)


Since the order matters (i.e., who gets gold, who gets silver and who gets bronze matters), the number of ways that 3 teams can be chosen from 10 teams is:

10P3 = 10 x 9 x 8 = (10 x 9 x 8 x 7!)/7! = 10!/7!

Answer: A
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Re: Medals are to be awarded to three teams in a 10-team [#permalink] New post   20 Jan 2021, 05:09
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There are 10 ways to select the first team. - which means 9 left.
So 9 ways to select 2nd team - which means 8 teams left
Finally, 8 ways to select the 3rd team

So no. of ways to select the set = 10x9x8

Option (a) 10!/7! can also be written as (10x9x8x7x6 ~ x1)/(7x6 ~x1) = 10x9x8. so its the right answer
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Re: Medals are to be awarded to three teams in a 10-team [#permalink] New post   12 Dec 2023, 00:16
If counting using the "boxes" method:
[1st box] = 10 options
[2nd box] = 9 options
[3rd box] = 8 options
All these shall be multiplied in accordance with the basic counting principle.

However, the options are expressed in a non-reduced form. So, the equivalent form in a fraction reduced that way so we have 10*9*8 in its nominator.
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Re: Medals are to be awarded to three teams in a 10-team [#permalink]
12 Dec 2023, 00:16
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