I have solved this in below fashion
When 'N' finishes second then there are 4 places where G can take (i.e 4 ways)
When 'N' finishes third then there are 3 places where G can take (i.e 3 ways)
No of ways is 5*4*3*2 = 120.
please explain flaw in my explanation.
Thanks in Advance
The most effective way to solve this is Bunuel's. However if we wanna take your approach more calculus are needed:
first of all 6 spots to fill _ _ _ _ _ _
If 'N' finishes first then the other 5 spots can be filled in \(5!\) ways N 5 4 3 2 1
And now it gets complicated...
If 'N' second then the other 5 spots can be filled in \(4*4!\) ways 4 N 4 3 2 1
. Why this?
N is in second position and there are 5 horses left. Of those only 4 can occupy the first position (every horse EXCEPT g), so write 4 on the first line.
We have 4 slots left and 4 horses => 4 3 2 1 for the remaining spots
With the same method if N finished 3rd, we get \(4*3*3!\) ways 4 3 N 3 2 1
N on the third slot, 5 horses left. The first spot can be occupied by 4 horses (every horse EXCEPT g), the second can be occupied by three horses (every horse EXCEPT g and the previusly chosen one); then for the others spots we have 3 horses => 3! ways.
if N finished 4th, we get \(4*3*2*2!\) ways 4 3 2 N 2 1
if N finished 5th, we get \(4*3*2*1\) ways 4 3 2 1 N 1
Sum them up and you get 360. This approach however is really long
Hope it's clear
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason
Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant
Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]