vishnuvardhan777 wrote:

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 1And 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 1N 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 1if N finished 5th, we get \(4*3*2*1\) ways

4 3 2 1 N 1Sum 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 QuantRules for Posting in the Verbal Forum -

Rules for Posting in the Quant Forum[/size][/color][/b]