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

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iwillcrackgmat
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Five people are running in a race. The first one to finish [#permalink] New post   13 Mar 2012, 20:15
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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
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C
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Bunuel
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Re: Five people are running in a race. The first one to finish [#permalink] New post   23 Mar 2012, 02:08
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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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subhashghosh
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Re: permutations & Combinations [#permalink] New post   13 Mar 2012, 20:34
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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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iwillcrackgmat
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Re: permutations & Combinations [#permalink] New post   13 Mar 2012, 21:14
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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subhashghosh
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Re: permutations & Combinations [#permalink] New post   13 Mar 2012, 21:19
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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pappueshwar
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Re: Five people are running in a race. The first one to finish [#permalink] New post   22 Mar 2012, 09:31
hey all,
whats the answer?

i got it as 60 . is that wrong ?
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SaraLotfy
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Re: Five people are running in a race. The first one to finish [#permalink] New post   18 Oct 2013, 01:15
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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Bunuel
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Re: Five people are running in a race. The first one to finish [#permalink] New post   18 Oct 2013, 05:37
Expert Reply
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.
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Re: Five people are running in a race. The first one to finish [#permalink] New post   17 Apr 2023, 09:02
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).

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Re: Five people are running in a race. The first one to finish [#permalink]
17 Apr 2023, 09:02
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