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# M and N are among the 5 runners in a race, and there can be no tie.

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Math Expert
Joined: 02 Sep 2009
Posts: 56366
M and N are among the 5 runners in a race, and there can be no tie.  [#permalink]

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01 Aug 2017, 12:16
10
00:00

Difficulty:

55% (hard)

Question Stats:

51% (01:20) correct 49% (01:32) wrong based on 118 sessions

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M and N are among the 5 runners in a race, and there can be no tie. How many possible results are there where M is ahead of N?

A. 10
B. 25
C. 35
D. 60
E. 80

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Joined: 18 Aug 2016
Posts: 617
Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29
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Re: M and N are among the 5 runners in a race, and there can be no tie.  [#permalink]

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01 Aug 2017, 21:34
8
1
Bunuel wrote:
M and N are among the 5 runners in a race, and there can be no tie. How many possible results are there where M is ahead of N?

A. 10
B. 25
C. 35
D. 60
E. 80

Arrangement of runners in 1st to 5th position = 5! = 120
M can either be ahead or behind N
Hence Probability of M ahead of N = 1/2 * 120 = 60

D
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Re: M and N are among the 5 runners in a race, and there can be no tie.  [#permalink]

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01 Aug 2017, 14:04
2
I'm not sure whether the question wants us to consider the positions of the remaining participants.

If the question doesn't, I guess the answer is A (10)

If it does, then the answer is D (60)

My explanation.

All the possible positions of M and N when M is ahead of N.

If N is in the 5th place, then M can be 4th, 3rd, 2nd and 1st. So 4 ways
If N is in the 4th place, then M can be 3rd, 2nd and 1st. 3 ways
If N is in the 3rd place, then M can be 2nd and 1st. 2 ways
And If N is in the 2nd place, then M only can be the 1st. 1 way

So in total, we have 4+3+2+1 = 10 ways, not considering the positions of the remaining 3 runners.

If we need to, then per each case, there are 3 free positions left for the 3 runners. So we need to allocate those positions among the 3 runners. We can do so 3!=6 ways in each of the possibilities above, regardless the places of M and N.
So multiplying each case by 6 will give us the answer.

(4+3+2+1)*6 = 60.

Please correct me If I'm wrong.
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Joined: 09 Sep 2013
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Re: M and N are among the 5 runners in a race, and there can be no tie.  [#permalink]

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25 Aug 2018, 12:46
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Re: M and N are among the 5 runners in a race, and there can be no tie.   [#permalink] 25 Aug 2018, 12:46
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# M and N are among the 5 runners in a race, and there can be no tie.

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