Five runners compete in a race for which first, second, and third plac

Question

Five runners compete in a race for which first, second, and third place prizes are awarded. In how many different ways could these prizes be awarded to these runners?

A. 5
B. 20
C. 60
D. 90
E. 120

leomanna10 · Answer

IMO E

G S B N N
5x4x3x2x1= 120

anmit123 · Answer

It should be 5x4x3=60

we don't care about 4th (and 5th position), we are concerned only with top 3.

GMATinsight · Answer

Method 1:

Three awards are to be given to three member so we can select three members out of 5 in 5C3 = 10 ways

Each group of selected members may be arranged on three distinct    Method 2:

The first award may be given to any one of the 5 members = 5 ways
The Second award may be given to any one of the remaining 4 members = 4 ways
The Third award may be given to any one of the remaining 3 members = 3 ways

Total ways to award three positions = 5*4*3 = 60 ways

Answer: Option C

Archit3110 · Answer

total players 5
prizes 3 
so prize can be arranged ; 5*4*3 = 60 ways
IMO C

Peachie · Answer

Here is how I solved mine, which I hope you will find relatively easy.
So we have 5 runners competing for 1st, 2nd and 3rd positions.
The 1st position will have 5 people competing for it, one person will win, 4 won't, so 5!/1!4!=5
Second position will have 4 people competing for it, so 1 person wins it, 3 won't, so 
4!/1!3=4
The 3rd position will have the remaining 3 people competing for it, 1 person will win it, 2 won't, so you have it as
3!/1!2!=3
Putting it all together 5*4*3=60 And is C

That is, 5!/1!4*4!/1!3!*3/1!2!=60
Really easy if you would apply it in this sense. 
You can try my metrhod in other combination problems and let me know how it helped you. I y sef to struggle woth combination problems before now as well, this was part of my breakthrough, can't even believe i' m teaching it now.
Kudos if my explanation cleared your confusion

Posted from my mobile device

Peachie · Answer

Here is how I solved mine, which I hope you will find relatively easy.
So we have 5 runners competing for 1st, 2nd and 3rd positions.
The 1st position will have 5 people competing for it, one person will win, 4 won't, so 5!/1!4!=5
Second position will have 4 people competing for it, so 1 person wins it, 3 won't, so 
4!/1!3=4
The 3rd position will have the remaining 3 people competing for it, 1 person will win it, 2 won't, so you have it as
3!/1!2!=3
Putting it all together 5*4*3=60 And is C

That is, 5!/1!4*4!/1!3!*3/1!2!=60
Really easy if you would apply it in this sense. 
You can try my metrhod in other combination problems and let me know how it helped you. I y sef to struggle woth combination problems before now as well, this was part of my breakthrough, can't even believe i' m teaching it now.
Kudos if my explanation cleared your confusion

Posted from my mobile device

ScottTargetTestPrep · Answer

This is a permutation problem because the order of finishing makes a difference. For example, (ABC) is a different ordering than (BAC). Thus, the prizes can be awarded in 5P3 = 5!/(5 - 3)! = 5!/2! = 5 x 4 x 3 = 60 ways.

Answer: C

bumpbot · Answer

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