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.
