from the Australian Intermediate Mathematics Olympiad:
Hello, delicate_tears!
Let $\displaystyle S,\,T\,,F$ = number of 6-, 10-, and 15-passenger taxis, respectively,
. . where: .$\displaystyle S,\,T,\,F\:\leq\:6$
We have: .$\displaystyle 6S + 10T + 15F \:=\:120$
. . Hence: .$\displaystyle F \:=\:\frac{120-6S - 10T}{15} \quad\Rightarrow\quad F \:=\;8\; -\; \frac{2S}{5}\;-\; \frac{2T}{3}$ .[1]
Since $\displaystyle F$ is an integer, .$\displaystyle \begin{array}{cc}S\text{ is a multiple of 5:} & S = 5h \\ T\text{ is a multiple of 3:} & T = 3k \end{array}$
. . Then [1] becomes: .$\displaystyle F \;=\;8 - 2h - 2k \;=\;8 - 2(h + k)$
Since $\displaystyle S,\,T\:\leq\:6$, then: .$\displaystyle h \:=\:0,1\text{ and }k \:=\:0,1,2$
It seems that there are: .$\displaystyle 2 \times 3 \:=\:6$ possible choices.
However, if $\displaystyle h = k = 0$, then $\displaystyle F = 8$ . . . not allowed.
Therefore, there are $\displaystyle \bf{\color{blue}5}$ ways.