parmenides51 wrote: Sofia has forgotten the passcode of her phone. She only remembers that it has four digits and that the product of its digits is $18$. How many passcodes satisfy these conditions?

Here

$a\cdot b \cdot c \cdot d=18$ Any permutations the cases below work $(1,1,2,9)$ $(1,1,3,6)$ $(1,3,3,2)$ Thus, our answer is $12 \cdot 3 = 36$

$abcd=18$. Factoring $18$, we have $3^2\cdot2$. Each of $a, b, c,$ and $d$ cannot contain over 2 factors lest the digit not be single-digit. By stars and bars/distributions, there are $\binom{5}{3}$ ways to distribute the two $3$'s to 4 values (i.e., $a, b, c, d$). There are also $\binom{4}{1}$ ways to distribute the 2 to the 4 values. Hence, there are 40 ways to do so. Subtracting the illegal distributions where one value has 3 factors (the two $3$'s and the one $1$) and the other digits are $1$. There are 4 configurations. Thus, there are a total of $40-4=\boxed{36}$ ways to do so.

@Grolarbear $40-4=36$

Ahahahaha I apologize - I simply do not comprehend basic arithmetic