Why does the combination formula divide counts by two?

The problem here is that $a$ and $c$ have two different roles. Therefore simply using $\binom{5}{2}$ will not work, as it denotes the number of ways of picking two elements out of five, but without distinguishing between them. What you can do is $$\binom{5}{2}\cdot2\cdot4 = 80\ ,$$ where we piked two numbers in $\{1,2,3,4,5\}$ using $\binom{5}{2}$, then we choose which one to assign to $a$ and which one to $c$ (whence the factor $2$), and finally we choose where to put the $c$ (the factor $4$).

There are $\binom{5}{2}$ ways to choose two numbers from the set of $5$. However, there are addditionally $2$ ways to choose which of those two numbers will be repeated three times. Given this, there are finally $4$ ways to choose which of the four positions the single non-repeated number will be in. This gives

$$\binom{5}{2} \cdot 2 \cdot 4 = \boxed{80\,}$$

This matches your answer using the other method.