Number of possible solutions for an equation 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?
 A: 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 $\binom{27}{2} = 351$.
A: 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.
A: HINT $\rm\ \ (x,y,z)\ \to\ \{x,\ x+y\}\ $ bijects solutions with two elt subsets of $\{1,2,\cdots,24\} $
A: 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.
