a 20x20 grid will be traversed in a lot of way. But each way is composed of 40 simple actions
_simple_action_DOWN: from current point move one step down (1 unit)
_simple_action_LEFT: from current point move one step left (1 unit)

So the question is equivalent to: given 40 indexes place 20 DOWN and 20 LEFT on them
Meaning given 40 indexex that are all DOWN , chose 20 random ones and palce LEFT on them. meaning Combinations of 40 taken by 20.

Same generalization for a rectangle of sizes L and l : given L+l position , choose randomly L of them = Combinations of (L+l) taken by l

Sorry for the English , not my native language.

Ryan McGuire
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Joined: Feb 18, 2005
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Ok, how about this variation:

How many ways are there to get from (0,0) to (20,20) (in the lower right quadrant, for the sake of discussion) if you're allowed to backtrack at most once? i.e. You're allowed one step to the left or one step up, but not both.

Steve Fahlbusch wrote:FYI: Myke's post should have been the best post as the answer is simply 20C40. And yes a py tri will work, but why. You don't need it.

I was trying to give a hint without giving away the entire solution. Just saying "the answer is 'how can you choose 20 out of 40 elements'?" doesn't help the person understand WHY it is the answer.