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Bunuel
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its choosing 3 teams from 10 teams when the order matters
so its 10P3.
the ans is A
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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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I thought the ten choose three formula was (n choose k)= n!/(k!(n-k)!)

Am I over thinking this...
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georgea
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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Agreed, A. I used slot method as order matters
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anilnandyala
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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anilnandyala
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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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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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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