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In a particular 100 meter race, 6 athletes are participating that are

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In a particular 100 meter race, 6 athletes are participating that are [#permalink]

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29 Oct 2011, 01:41
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In a particular 100 meter race, 6 athletes are participating that are numbered 1-6. In how many ways athlete #5 can always finish ahead of #2 provided there's no tie among the athletes.

a) 120
b) 300
c) 360
d) 720
e) 1440

[Reveal] Spoiler:
apparently there's a formula to solve this $$\frac{n!}{2}$$. does anyone know the logic behind this formula?
[Reveal] Spoiler: OA

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Re: In a particular 100 meter race, 6 athletes are participating that are [#permalink]

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29 Oct 2011, 04:56
Consider the arrangement of the athletes: 1 2 3 4 5 6
The athletes can be considered as numbers to be arranged in 6 available slots: _ _ _ _ _ _
The first slot can be filled in 6 ways, the second slot in 5 ways and so on.
So, the total number of ways in which we can have the athletes in the slots is 6*5*4*3*2*1 = 6! = 720

Now, due to the symmetry of the arrangement, there can be exactly half the number of ways in which any one of the athletes can be above any other athlete.

This can be explained using probability. Consider the 6 slots: _ _ _ _ _ _
If we fill the slots randomly with the numbers 1 to 6, what is the probability of finding number 2 in the first slot? It would be 1/6. Similarly, what is the probability of finding the number 2 in the second slot? It is 1/6 again. And similarly, the probability of finding number 2 in the third slot is 1/6.
Hence the probability of finding the number 2 in the first OR second OR third slot is 1/6 + 1/6 + 1/6 = 3/6 = 1/2
In any of these conditions, the number 5 can occupy the fourth OR fifth OR sixth slot and hence the number 2 can be ahead or to the left of number 5 in half of the conditions or arrangement.

We apply this logic to the given question in order to get the number of possible results for the 100 meter race in which #5 finishes ahead of #2 given that there's no ties.
So in exactly half of the possible arrangements i.e. 6!/2 i.e. 720/2 = 360 results will be such that #5 finishes ahead of #2 in the race.

This logic can be applied in questions with symmetrical arrangements such as seating people, creating ranklists, etc.
Hope this helps.
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Re: In a particular 100 meter race, 6 athletes are participating that are [#permalink]

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29 Oct 2011, 05:49
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Ans is 360 ways.

Case1: 5 _ _ _ _ _
#5 is at first place. The remaining 5 places can be filled in 5! ways= 120

Case 2: _5_ _ _ _
#5 is at second place. Ways of filling first place will be 4 as #2 cannot be there so 4*1*4*3*2*1 = 96

Case 3: _ _ 5 _ _ _
#5 is at third place. Ways of filling first and second places will be 4 and 3 respectively as 2 cannot be before 5. So 4*3*1*3*2*1 = 72

Case 4: _ _ _ 5 _ _
#5 is at fourth place. Ways of filling first, second and third places will be 4*3*2. Total ways = 4*3*2*1*2*1= 48

Case 5: _ _ _ _ 5 _
#5 is at fifth place. Ways of filiing first, second, third and fourth places will be 4* 3* 2* 1 * 1* 1= 24

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Re: In a particular 100 meter race, 6 athletes are participating that are [#permalink]

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30 Oct 2011, 11:45
@ashimasood - nice explanation. Kudos to you.

Cheers!
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Re: In a particular 100 meter race, 6 athletes are participating that are [#permalink]

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30 Aug 2016, 08:02
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Re: In a particular 100 meter race, 6 athletes are participating that are [#permalink]

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17 Sep 2016, 15:39
After a lot of thinking, below is my reasoning: Arrangements below are last to first (left to right)

Below are the possibilities:

Case a)
2 _ _ _ _ _ < #5 can be placed in any of the 5 available spots before #2 so there are 5 options. After placing #5 in any of these 5 spots, remaining athletes can be arranged in 4! ways. So total combinations for this case is 5*4!

Case b)
_ 2 _ _ _ _ < #5 can be placed in any of the 4 available spots before #2 so there are 4 options. After placing #5 in any of these 4 spots, remaining athletes can be arranged in 4! ways. So total combinations for this case is 4*4!

Case c)
_ _ 2 _ _ _ < #5 can be placed in any of the 3 available spots before #2 so there are 3 options. After placing #5 in any of these 3 spots, remaining athletes can be arranged in 4! ways. So total combinations for this case is 3*4!

Case d)
_ _ _ 2 _ _ < #5 can be placed in any of the 2 available spots before #2 so there are 2 options. After placed 5 in any of these 2 spots, remaining athletes can be arranged in 4! ways. So total combinations for this case is 2*4!

Case e)
_ _ _ _ 2 _ < #5 can be placed in only one spot for this case. Remaining athletes can be arranged in 4! ways. So total combinations for this case is 1*4!

Total combinations = Case a + b + c + d + e = 4!(5+4+3+2+1) = 4!*15 = 24*15 = 360.

I guess this is similar to ashimasood's method
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In a particular 100 meter race, 6 athletes are participating that are [#permalink]

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18 Sep 2016, 22:28
First step: calculate all the possible arrangements: 6! = 720.

There is 50% chance that a certain runner will finish before his competitor.

720*0.5 = 360
In a particular 100 meter race, 6 athletes are participating that are   [#permalink] 18 Sep 2016, 22:28
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