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They are different.

woops!! that's for ... a+b+c+d = N

a+b = n would make n+1 solutions
a +b + c = n would make
|dw:1350155772870:dw|

probably we would add up (summation) n-2 times for n+1

Hm.. I would like you to write the whole solution and the expression as an answer.

honestly I don't know any closed form for this.

who are you taking to? @experimentX or me?

You both. I know the answer. I found this problem very interesting.

you just need to generalize...

well ... one thing is for sure ... it must have something with to do of the order of N^(N+1)

You have n units and (N-1) s's. How many ways can you select (N-1) things out of (n-1)

Think again. In an hour I will give you an answer.

1:15 am here ...

Do you want it now?

I'll wait till tomorrow.

Right, the are are a few solutions missing - the ones that start with 0 or end in 0

probably this is related to this
http://en.wikipedia.org/wiki/Partition_(number_theory)

why not just add s's at the beginning and end of the sequence...

never mind...that won't take care of the others...

can x's be zero?

yes, it can, as can any of the variables. But my take on the problem excludes those possibilities.

My solution won't allow for adjacent s's

Claim: every arrangement of n u's and (N-1) s's gives a solution.

so you have a total of n + (N-1) things and you choose where to place (N-1) of them:
C(n+N-1, N-1)

so C(n+N-1, N-1) is your solution?

|dw:1350157242721:dw|

|dw:1350157497037:dw|

interesting!!

Did you understand or not?

yeah i see the picture ...

Very good!

never thought i would have such interestingly simple technique.

*it

Try to solve my next ploblem using similar method.

sure ..