5 apples are randomly distributed to 4 boxes. We need to find probability that there are 2 boxes with 2 apples, 1 box with 1 apple and 1 empty box.

I'm getting the correct answer with $\frac{\frac{5!}{2!2!1!0!}\ast 4\ast 3}{{4}^{5}}=0.3515625$ (anyway, the answer is said to be 0.35, but I think it is a matter of rounding).

But I don't understand why there are ${4}^{5}$ elementary events in total. Firstly, I thought It should be $(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\textstyle (}\genfrac{}{}{0ex}{}{4}{5}{\textstyle )}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}})$ - number of combinations with repetitions, but I couldn't get the proper answer.

Isn't approach with $(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\textstyle (}\genfrac{}{}{0ex}{}{4}{5}{\textstyle )}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}})$ elementary events more correct?