Solution 1:
There can be 6! different possibilities alltogether. Now half of them will be where GoldenRod will be in first 3 and No hope in second 3 i.e. in half of them, one will be ahead of another.
So total combinations is 360.
Solution 2: ( Found on internet)
if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! ways, so total ways 5*4!
if A finishes 2... B could be in any of the last 4 positions in 4 ways. but the other positions could be filled in 4! ways, so the total ways 4*4!
if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!
if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4!
if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..
A cannot finish 6th, since he has to be ahead of B
therefore total number of ways
5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360
I dont think either of them are the best approches..Can anyone provide a better and simpler approach ?
Thanks
Airan
