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# Four contestants representing four different countries advance to the

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Math Expert
Joined: 02 Sep 2009
Posts: 54544

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08 Dec 2014, 06:21
00:00

Difficulty:

15% (low)

Question Stats:

73% (00:57) correct 27% (00:59) wrong based on 172 sessions

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Tough and Tricky questions: Combinations.

Four contestants representing four different countries advance to the finals of a fencing championship. Assuming all competitors have an equal chance of winning, how many possibilities are there with respect to how a first-place and second-place medal can be awarded?

A) 6
B) 7
C) 12
D) 16
E) 24

Kudos for a correct solution.

Source: Chili Hot GMAT

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Joined: 04 Oct 2013
Posts: 151
Location: India
GMAT Date: 05-23-2015
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08 Dec 2014, 08:24
1
Four contestants representing four different countries advance to the finals of a fencing championship. Assuming all competitors have an equal chance of winning, how many possibilities are there with respect to how a first-place and second-place medal can be awarded?

A) 6
B) 7
C) 12
D) 16
E) 24

Number of ways First-place medal can be awarded to four contestants = 4
Number of ways Second-place medal can be awarded to contestants after awarding First-place medal =3

Therefore number of possibilities = 4 *3 =12
Intern
Joined: 08 Jul 2012
Posts: 46

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08 Dec 2014, 21:11
1
Four contestants representing four different countries advance to the finals of a fencing championship. Assuming all competitors have an equal chance of winning, how many possibilities are there with respect to how a first-place and second-place medal can be awarded?

We have 2 slots to be filled using 4 contestants: 4 options for slot1 * 3 option for slot2
= 4* 3
= 12

Ans. C) 12
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Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time. - Thomas A. Edison
Director
Joined: 23 Jan 2013
Posts: 549
Schools: Cambridge'16

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08 Dec 2014, 22:16
1
As I see we talk about how many possibilities of placing 4 distinct objects
with regard to 1 and 2 place

All possibilities: 4!=24

Restrictions: two places regard and two not regard multiplyed by 2 so, (4!/2!*2!)*2=12

C
Math Expert
Joined: 02 Sep 2009
Posts: 54544

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09 Dec 2014, 06:46
Bunuel wrote:

Tough and Tricky questions: Combinations.

Four contestants representing four different countries advance to the finals of a fencing championship. Assuming all competitors have an equal chance of winning, how many possibilities are there with respect to how a first-place and second-place medal can be awarded?

A) 6
B) 7
C) 12
D) 16
E) 24

Kudos for a correct solution.

Source: Chili Hot GMAT

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Joined: 07 Oct 2014
Posts: 12

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14 Dec 2014, 15:10
isnt 4C2 (i.e 6 ways) the answer ?

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 54544

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15 Dec 2014, 07:02
1
vinbitstarter wrote:
Four contestants representing four different countries advance to the finals of a fencing championship. Assuming all competitors have an equal chance of winning, how many possibilities are there with respect to how a first-place and second-place medal can be awarded?

A) 6
B) 7
C) 12
D) 16
E) 24

isnt 4C2 (i.e 6 ways) the answer ?

Thanks

No. 4C2 gives the number of two-contestant groups possible from 4. Suppose we are choosing group {A, B}. We can have A = first-place and B = second-place or vise-versa, so for each group there are 2 possible arrangement with respect to how a first-place and second-place medal can be awarded. Therefore the answer is 4C2*2 = 12.

Hope it's clear.
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Joined: 07 Oct 2014
Posts: 12

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20 Dec 2014, 21:57
Bunuel wrote:
vinbitstarter wrote:
Four contestants representing four different countries advance to the finals of a fencing championship. Assuming all competitors have an equal chance of winning, how many possibilities are there with respect to how a first-place and second-place medal can be awarded?

A) 6
B) 7
C) 12
D) 16
E) 24

isnt 4C2 (i.e 6 ways) the answer ?

Thanks

No. 4C2 gives the number of two-contestant groups possible from 4. Suppose we are choosing group {A, B}. We can have A = first-place and B = second-place or vise-versa, so for each group there are 2 possible arrangement with respect to how a first-place and second-place medal can be awarded. Therefore the answer is 4C2*2 = 12.

Hope it's clear.

Yes, thank you
another way to look at it is : 4C1 * 3C1 = 12 possible ways.
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Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 413
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)

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26 Dec 2014, 02:39
Hi,

Let me share the 2 ways in which I did it. Please, let me know if there is a mistake somewhere in my thinking or a possible trap in these 2 solutions:

1st solution:
4!=24 possible ways
24/2=12, because it is the first 2 spots that will be changing. It is sort of intuitive, so it could easily be wrong...

I tried the 2nd solution solution, just to see if I end up with 12 again:
We have A - B - C - D as the possible slots. The first 2 slots can change so that it is occupied by 2 different people.
Starting with A we have:
A - B - C - D
A - C - B - D
A - D - B - C
So, these are 3 ways, only taking into account A. We gave 4 different people, so 3 X 4 = 12 (just thinking that each one of them should have the same chances).
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04 Sep 2017, 08:11
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Re: Four contestants representing four different countries advance to the   [#permalink] 04 Sep 2017, 08:11
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