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If $x$, $y$, and $z$ are natural numbers, how many solutions are there to $x+y+z=25$?

How would I figure this out? I can't even begin dissecting this problem. Where do I begin?

4 Answers 4

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Suppose that "natural number" doesn't include 0. Take 25 balls, and put a wall between any two. A partition $a+b+c=25$ is the same as a choice of two walls, and there are $\binom{24}{2} = 276$ of these.

If zeroes are allowed, then $(x+1)+(y+1)+(z+1) = 28$, and so the answer is $\binom{27}{2} = 351$.

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    Lol oh come on man..I dont see it.2010-11-08
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This is called the "partition function" of 25. See here:

http://en.wikipedia.org/wiki/Partition_function_%28number_theory%29

It's not an easy-to-compute-directly function - the best bet is simply writing a small script that counts solutions using a double loop.

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    This [suggested edit](http://math.stacke$x$change.com/review/suggested-edits/149189) was probably supposed to be a comment.2014-01-10
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HINT $\rm\ \ (x,y,z)\ \to\ \{x,\ x+y\}\ $ bijects solutions with two elt subsets of $\{1,2,\cdots,24\} $

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Here is a simpler explanation: Lets look at the equation x + y = k where x >= 0 && y >=0 the solution set is of form {{k, 0}, {k-1, 1}, ... {0, k+1}} and there are (k+1) items in the set.

x + y + z = k can be broken down into: x + (y + z) = k

if we choose x = 0, solutions of (y + z) = k are k+1. if we choose x = 1, solutions of (y + z) = k-1 are k. if we choose x = 2, solutions of (y + z) = k-2 are k-1. if we choose x = k, solutions of (y + z) = 0 are 1.

this is a simple natural number sum: 1 + 2 + 3 + ... + (k +1) = ((k + 1) * (k +2)/ 2)

My math is weak so I dont know what are the proper names to call something.