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Four contestants representing four different countries advance to the

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Four contestants representing four different countries advance to the [#permalink]

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New post 08 Dec 2014, 06:21
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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
[Reveal] Spoiler: OA

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Re: Four contestants representing four different countries advance to the [#permalink]

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New post 08 Dec 2014, 08:24
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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
Answer: C

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New post 08 Dec 2014, 21:11
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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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New post 08 Dec 2014, 22:16
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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

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Re: Four contestants representing four different countries advance to the [#permalink]

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New post 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


The correct answer is C.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

Thanks

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New post 15 Dec 2014, 07:02
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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 ?
please clarify

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.

Answer: C.

Hope it's clear.
_________________

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Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 129030 [1], given: 12187

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Re: Four contestants representing four different countries advance to the [#permalink]

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New post 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 ?
please clarify

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.

Answer: C.

Hope it's clear.



Yes, thank you
another way to look at it is : 4C1 * 3C1 = 12 possible ways.

Kudos [?]: [0], given: 12

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Re: Four contestants representing four different countries advance to the [#permalink]

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New post 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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Re: Four contestants representing four different countries advance to the   [#permalink] 04 Sep 2017, 08:11
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