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Five people are running in a race. The first one to finish [#permalink]
13 Mar 2012, 19:15
1
This post received KUDOS
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00:00
A
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Difficulty:
15% (low)
Question Stats:
70% (01:33) correct
30% (00:36) wrong based on 148 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?
Re: permutations & Combinations [#permalink]
13 Mar 2012, 20:14
2
This post received KUDOS
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? _________________
Press +1 Kudos rather than saying thanks which is more helpful infact..
Re: Five people are running in a race. The first one to finish [#permalink]
23 Mar 2012, 01:08
2
This post received KUDOS
Expert's post
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This post was BOOKMARKED
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 \(P^3_5=60\)).
Re: Five people are running in a race. The first one to finish [#permalink]
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? _________________
Re: Five people are running in a race. The first one to finish [#permalink]
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
Re: Five people are running in a race. The first one to finish [#permalink]
30 Oct 2014, 08:20
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