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klimenkov
 4 years ago
How many nonnegative integer solutions does this equation have?
\[x_1+x_2+\ldots+x_N=n\]
klimenkov
 4 years ago
How many nonnegative integer solutions does this equation have? \[x_1+x_2+\ldots+x_N=n\]

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Are 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.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0woops!! that's for ... a+b+c+d = N

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For 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.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0a+b = n would make n+1 solutions a +b + c = n would make dw:1350155772870:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0probably we would add up (summation) n2 times for n+1

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0Hm.. I would like you to write the whole solution and the expression as an answer.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0honestly I don't know any closed form for this.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0who are you taking to? @experimentX or me?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For 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.

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0You both. I know the answer. I found this problem very interesting.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you just need to generalize...

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0well ... one thing is for sure ... it must have something with to do of the order of N^(N+1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You have n units and (N1) s's. How many ways can you select (N1) things out of (n1)

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0Think again. In an hour I will give you an answer.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0I'll wait till tomorrow.

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0@cruffo 's method is very good. But he must correct it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Right, the are are a few solutions missing  the ones that start with 0 or end in 0

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0probably this is related to this http://en.wikipedia.org/wiki/Partition_(number_theory)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why not just add s's at the beginning and end of the sequence...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0never mind...that won't take care of the others...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, it can, as can any of the variables. But my take on the problem excludes those possibilities.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My solution won't allow for adjacent s's

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Claim: every arrangement of n u's and (N1) s's gives a solution.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so you have a total of n + (N1) things and you choose where to place (N1) of them: C(n+N1, N1)

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0so C(n+N1, N1) is your solution?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1350157242721:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1350157497037:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0can't think at the moment ... Night!! will see solution tomorrow directly. Nevertheless combinatorics has never been my stuff.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"Mathematics of Choice" by Ivan Niven http://books.google.com/books?id=UrVQYgEACAAJ The best introductory book, IMHO

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0@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 \((n+N+1)2=n+N1\) 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 \(\left(\begin{matrix}n+N1 \\ N1\end{matrix}\right)\) ways because we can shift \(N1\) 1's. Or we can shift 0's. then we will have \(\left(\begin{matrix}n+N1 \\ n\end{matrix}\right)\) ways. This will be an answer. Hope you understood and enjoyed this interesting problem.

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0Did you understand or not?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0yeah i see the picture ...

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0never thought i would have such interestingly simple technique.

klimenkov
 4 years ago
Best ResponseYou've already chosen the best response.0Try to solve my next ploblem using similar method.
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