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How many non-negative integer solutions does this equation have? \[x_1+x_2+\ldots+x_N=n\]

Mathematics
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Are you considering the solutions to be ordered or unordered? For example, if we wee trying to solve: x + y + z + w = 12 Would the solution (x,y,z,w) = (9,1,1,1) = (1,9,1,1) = (1,1,9,1) = (1,1,1,9) or would that be considered 4 separate solutions? If they are considered the same, then this is a partitions problem, if they are different, then this is a combinations problem.
They are different.
woops!! that's for ... a+b+c+d = N

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Other answers:

For the smaller example: x + y + z + w = 12 The number of solutions can be determined by viewing the problem in the following way: If 12 units (denoted as 12 u's) seperated by 11 spaces (denoted by 11 s's) are lined up, usususususususususususu and we choose any 3 of the s's and let the others disappear, for example: u u s u u u usu u u u usu then the remaining s's seperate the units into four batches. The number of units in these batches can be used as the values of x,y,z, and w.
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?
For mine, any selection of three s's gives a solution. Thus, the number of solutions is the same as the number of ways of selection 3 things out of 11.
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.
@cruffo 's method is very good. But he must correct it.
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|
can't think at the moment ... Night!! will see solution tomorrow directly. Nevertheless combinatorics has never been my stuff.
"Mathematics of Choice" by Ivan Niven http://books.google.com/books?id=UrVQYgEACAAJ The best introductory book, IMHO
@cruffo is right. I will write a full solution. Let's make a simple model of this problem. Imagine that we have \(n\) balls. There is no matter what is first, what is the second,..., they are all the same. And we have \(N\) boxes where to put this balls. You can put this \(n\) balls in any box that you like. On the picture there are some ways how to put 10 balls into 5 boxes. The number of the balls in the first box is \(x_1\), in the second - \(x_2\) ... in the \(N\)-th - \(x_N\).|dw:1350209158710:dw| Lets make the side of the box as 1 and about the ball as 0. Every placement of the balls in this boxes can be shown as the sequence of \(n+N+1\) zeros and 1's. For \(n=5\) and \(N=10\) an example will be 1000011000011001. But the first and the last term in this sequence must be 1 because they are the sides of the box. If we shift 0's and 1's inside the first and the last 1's we will get the new arrangement in the boxes. But we can replace only \((n+N+1)-2=n+N-1\) terms in the sequence. If we choose the places for 1's the place for 0's will be chosen automatically. So we can choose the place for 1's in \(\left(\begin{matrix}n+N-1 \\ N-1\end{matrix}\right)\) ways because we can shift \(N-1\) 1's. Or we can shift 0's. then we will have \(\left(\begin{matrix}n+N-1 \\ n\end{matrix}\right)\) ways. This will be an answer. Hope you understood and enjoyed this interesting problem.
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 ..

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