# 2005 Alabama ARML TST Problems/Problem 12

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## Problem

Find the number of ordered pairs of positive integers $(a,b,c,d)$ that satisfy the following equation:

$a+b+c+d=12$.

## Solution

### Solution 1

The generating function for $a, b, c,$ and $d$ is $x+x^2+x^3+\cdots$.

$$(x+x^2+x^3+\cdots)^4=x^4(1+x+x^2+x^3)^4= x^4+{4\choose 3}x^5+{5\choose 3}x^6+\cdots$$

The coefficient of $x^{12}$ is $\binom{3+12-4}{3}=165$.

### Solution 2

We consider a bijection to the following combinatorial argument. Suppose we have twelve balls and we would like to put them into four urns. There is a positive number of balls in each urn, so we put one ball into each urn initially. We are left with eight balls to go into four urns, which is equivalent to have three dividers. Thus there are $\frac{11!}{8!3!} = 165$ ordered pairs.

## See also

 2005 Alabama ARML TST (Problems) Preceded by:Problem 11 Followed by:Problem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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