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Five people are running in a race. The first one to finish

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Manager
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Five people are running in a race. The first one to finish [#permalink] New post 13 Mar 2012, 19:15
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A
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C
D
E

Difficulty:

  15% (low)

Question Stats:

70% (01:32) correct 30% (00:36) wrong based on 142 sessions
Five people are running in a race. The first one to finish wins a gold medal, the second wins a silver medal and the third wins a bronze medal. How many different arrangements of medal winners, in order from first to third, are possible?

A. 5
B. 10
C. 60
D. 120
E. 125
[Reveal] Spoiler: OA

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Re: permutations & Combinations [#permalink] New post 13 Mar 2012, 19:34
Assuming no ties in the race :

The Gold Medal can be given in 5 ways
The Silver Medal can be given in 4 ways
The Bronze Medal can be given in 3 ways

So total number of ways = 5 * 4 * 3 = 60

Answer - C
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2 KUDOS received
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Joined: 22 Jan 2012
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Re: permutations & Combinations [#permalink] New post 13 Mar 2012, 20:14
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same type of question where order doesn't matter

Five people are running in a race. The first three to finish win
gift certificates. How many different groups of people could
win the gift certificates?
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Re: permutations & Combinations [#permalink] New post 13 Mar 2012, 20:19
Here it's 5C3 = 5!/3!2!
= 5*4/2 = 10

The idea is to pick a group of 3 people out of 5.
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Re: Five people are running in a race. The first one to finish [#permalink] New post 22 Mar 2012, 08:31
hey all,
whats the answer?

i got it as 60 . is that wrong ?
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Re: Five people are running in a race. The first one to finish [#permalink] New post 23 Mar 2012, 01:08
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pappueshwar wrote:
hey all,
whats the answer?

i got it as 60 . is that wrong ?


OA is given under the spoiler in the initial post.

Five people are running in a race. The first one to finish wins a gold medal, the second wins a silver medal and the third wins a bronze medal. How many different arrangements of medal winners, in order from first to third, are possible?
A. 5
B. 10
C. 60
D. 120
E. 125

\(C^3_5*3!=60\), where \(C^3_5\) is ways to select the group of 3 winners out of 5 contestants and 3! is ways to arranging them (alternately you can just do \(P^3_5=60\)).

Answer: C.

Hope it helps.
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Re: Five people are running in a race. The first one to finish [#permalink] New post 18 Oct 2013, 00:15
Bunuel wrote:
pappueshwar wrote:
hey all,
whats the answer?

i got it as 60 . is that wrong ?


OA is given under the spoiler in the initial post.

Five people are running in a race. The first one to finish wins a gold medal, the second wins a silver medal and the third wins a bronze medal. How many different arrangements of medal winners, in order from first to third, are possible?
A. 5
B. 10
C. 60
D. 120
E. 125

\(C^3_5*3!=60\), where \(C^3_5\) is ways to select the group of 3 winners out of 5 contestants and 3! is ways to arranging them (alternately you can just do \(P^3_5=60\)).

Answer: C.

Hope it helps.



Hello Bunuel,

I didn't understand why you used \(P^3_5=60\) to solve this question. Also, the actual evaluation of it seems to get me - I know this is a silly point but I still get it wrong -

I evaluate as follows:

\(P^3_5\)= 5!/3! = 5*4*3*2*1/ 3*2*1 = 20. where am I going wrong?
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Re: Five people are running in a race. The first one to finish [#permalink] New post 18 Oct 2013, 04:37
Expert's post
SaraLotfy wrote:
Bunuel wrote:
pappueshwar wrote:
hey all,
whats the answer?

i got it as 60 . is that wrong ?


OA is given under the spoiler in the initial post.

Five people are running in a race. The first one to finish wins a gold medal, the second wins a silver medal and the third wins a bronze medal. How many different arrangements of medal winners, in order from first to third, are possible?
A. 5
B. 10
C. 60
D. 120
E. 125

\(C^3_5*3!=60\), where \(C^3_5\) is ways to select the group of 3 winners out of 5 contestants and 3! is ways to arranging them (alternately you can just do \(P^3_5=60\)).

Answer: C.

Hope it helps.



Hello Bunuel,

I didn't understand why you used \(P^3_5=60\) to solve this question. Also, the actual evaluation of it seems to get me - I know this is a silly point but I still get it wrong -

I evaluate as follows:

\(P^3_5\)= 5!/3! = 5*4*3*2*1/ 3*2*1 = 20. where am I going wrong?


First of all \(P^n_k=\frac{n!}{(n-k)!}\), thus \(P^3_5=5!/(5-3)!=60\).

Next, consider the runners to be A, B, C, D and E.

\(P^3_5\) gives all different ordered triplets from 5:
ABC
ACB
BAC
BCA
CAB
CBA

ABD
ADB
BAD
BDA
DAB
DBA
...

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis ; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) ; 12. Tricky questions from previous years.

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.


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Re: Five people are running in a race. The first one to finish [#permalink] New post 30 Oct 2014, 08:20
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Re: Five people are running in a race. The first one to finish   [#permalink] 30 Oct 2014, 08:20
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