Problem: An ordered pair $(m,n)$ of non-negative integers is called "simple" if the addition $m+n$ in base $10$ requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to $1492$. Solution: We can use a counting argument. Consider the four-digit number $a_1a_2a_3a_4$ where each $a_i$ is a digit. If we write $a_1a_2a_3a_4 = b_1b_2b_3b_4 + c_1c_2c_3c_4$, we must have that each $a_i = b_i+c_i$ so this is a partition problem. For each digit, $k$ we may write it as $k+1$ sums $0+k, 1+(k-1), 2+(k-2), \ldots, k+0$. Note the order matters since we are counting the number of simple ordered pairs, so the answer is $2 \cdot 5 \cdot 10 \cdot 3 = 300$.