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Re: general question

by Vemuri » Sat Apr 18, 2009 8:10 pm
ketkoag wrote:M and N are among the 5 runners for a race, there is no tie. How many possible results are there that M is ahead of N?

OA after few posts..
If M comes first, then N can finish the race in 4 ways.
If M comes second, then N can finish the race in 3 ways
If M comes third, then N can finish the race in 2 ways
If M comes fourth, then N can finish the race in 1 way

So, in all there are 4*3*2*1 = 24 ways.
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by dmateer25 » Sat Apr 18, 2009 8:56 pm
5! total results = 120

Out of the 120 M will be ahead of N half of the time.

120/2 = 60


Another way to look at it. Here are the possible ways for M to be ahead of N

MN_ _ _
M _ N _ _
M _ _ N _
M _ _ _ N
_ M N _ _
_ M _ N _
_ M _ _ N
_ _ M N _
_ _ M _ N
_ _ _ M N

For each possibility, the other 3 racers can finish in 3! ways.

so 3! * 10 = 60
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by ketkoag » Sun Apr 19, 2009 7:21 am
dmateer25 wrote:5! total results = 120

Out of the 120 M will be ahead of N half of the time.

120/2 = 60


Another way to look at it. Here are the possible ways for M to be ahead of N

MN_ _ _
M _ N _ _
M _ _ N _
M _ _ _ N
_ M N _ _
_ M _ N _
_ M _ _ N
_ _ M N _
_ _ M _ N
_ _ _ M N

For each possibility, the other 3 racers can finish in 3! ways.

so 3! * 10 = 60
Hi, OA is 60
please lemme know why M will be ahead of N half of the time.
please explain as i got the answer from the second method. please elaborate some more on the first method. :)
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by dmateer25 » Sun Apr 19, 2009 7:38 am
ketkoag wrote:
dmateer25 wrote:5! total results = 120

Out of the 120 M will be ahead of N half of the time.

120/2 = 60


Another way to look at it. Here are the possible ways for M to be ahead of N

MN_ _ _
M _ N _ _
M _ _ N _
M _ _ _ N
_ M N _ _
_ M _ N _
_ M _ _ N
_ _ M N _
_ _ M _ N
_ _ _ M N

For each possibility, the other 3 racers can finish in 3! ways.

so 3! * 10 = 60
Hi, OA is 60
please lemme know why M will be ahead of N half of the time.
please explain as i got the answer from the second method. please elaborate some more on the first method. :)
Sure.

If you think of the problem using my second method you will understand why M will be ahead of N half of the time.

MN_ _ _
M _ N _ _
M _ _ N _
M _ _ _ N
_ M N _ _
_ M _ N _
_ M _ _ N
_ _ M N _
_ _ M _ N
_ _ _ M N

So those are all the possibilities with M ahead of N.

Now, you switch the M and N and you have all the ways in which N will be ahead of M.

10 ways x 3! ways m is ahead of n = 60

10 ways x 3! ways n is ahead of m = 60

so you see, half of the possibilities, M finishes ahead of N and half of the possibilities, N finishes ahead of N.

Now if you understand that concept, you can use my first method.

There are 5! total outcomes = 120

Now because we know M will be ahead of N half of the time, just divide the 120 by 2 and you come up with 60.
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by lilu » Sun Apr 19, 2009 9:39 am
Thank you, dmateer25! That was a great explanation!
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