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

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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?

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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Press +1 Kudos rather than saying thanks which is more helpful infact..

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 \(P^3_5=60\)).

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

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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?
_________________

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

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

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30 Oct 2014, 09:20

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]

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27 May 2016, 23:41

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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