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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]

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13 Mar 2012, 19:15
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15% (low)

Question Stats:

72% (01:35) correct 28% (00:39) wrong based on 178 sessions

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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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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

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Manager
Joined: 22 Jan 2012
Posts: 90
Location: India
Concentration: General Management, Technology
GMAT 1: Q39 V29
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WE: Engineering (Consulting)
Followers: 6

Kudos [?]: 102 [2] , given: 9

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13 Mar 2012, 20:14
2
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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
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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]

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22 Mar 2012, 08:31
hey all,

i got it as 60 . is that wrong ?
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eshwar

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Re: Five people are running in a race. The first one to finish [#permalink]

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23 Mar 2012, 01:08
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Expert's post
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pappueshwar wrote:
hey all,

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 $$P^3_5=60$$).

Hope it helps.
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Re: Five people are running in a race. The first one to finish [#permalink]

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18 Oct 2013, 00:15
Bunuel wrote:
pappueshwar wrote:
hey all,

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 $$P^3_5=60$$).

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]

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18 Oct 2013, 04:37
SaraLotfy wrote:
Bunuel wrote:
pappueshwar wrote:
hey all,

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 $$P^3_5=60$$).

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
BDA
DAB
DBA
...

Hope it helps.
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Re: Five people are running in a race. The first one to finish [#permalink]

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

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27 May 2016, 22:41
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: Five people are running in a race. The first one to finish   [#permalink] 27 May 2016, 22:41
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