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