Four balls are randomly dropped into four boxes Four balls are randomly dropped into four boxes, where any ball is equally likely to fall into
each box.

For a fixed $k = 0, 1, 2, 3$, let $A_k$ denote the event that exactly $k$ boxes are empty. Compute
$P(A_k)$ for each $k = 0, 1, 2, 3$.

I have computed $P(A_0)$ and $P(A_3)$ which are $\frac{4!}{4^4}=\frac{3}{32}$ and $\frac{4}{4^4}=\frac{1}{64}$ recpectively. But I don't know how to find $P(A_1)$ and $P(A_2)$. Could anybody help me?
Imagine temporarily that each ball had its own color.  We see then that there are $4^4$ equally likely outcomes for where the balls are positioned.  We get as you already did the probabilities $P(A_0)=\dfrac{4!}{4^4}$ and $P(A_3)=\dfrac{4}{4^4}$

Now., for $P(A_1)$, count the numerator using the following multiplication principle argument:


Pick which box was empty: $4$ choices
Pick which box received two balls: $3$ choices
Pick which two colored balls go into the box designated to receive two balls: $\binom{4}{2}$ choices
Pick which colored ball goes into the left-most box designated to receive one ball: $2$ choices
The final ball goes into the final box designated to receive one ball


This gives a total of $4\cdot 3\cdot \binom{4}{2}\cdot 2$ options here for a probability then of

$$P(A_1)=\dfrac{144}{4^4}$$

We can similarly calculate for the other case, taking advantage of the fact that the boxes are arranged from left to right, giving a total of $\binom{4}{2}\binom{4}{2} = 36$ if there were two boxes, each with two balls in them and $4\cdot 3\cdot 4=48$ if there was one box with three and another box with only one, for a total of $84$ outcomes and a probability of:

$$P(A_2)=\dfrac{84}{4^4}$$

Checking, $4!+144+84+4 = 256$ and so we do get our probabilities adding to one as expected.
$P(A_0)=\frac{1}{35}$ because the only possibility is the every has exactly one ball.
$P(A_3)=\frac{4}{35}$ because there are 4 possibility to choose the one who isn't empty.
$P(A_1)=\frac{\binom{4}{1}\cdot3}{35}$ because the first binomial is the way to choose the one without the ball, and the way to give 4 balls to 3 people so that every of them has at least 1 ball.
$P(A_2)=\frac{\binom{4}{2}\cdot\binom{3}{2}}{35}$ because the first binomial is the way to choose the two without the ball, and the second binomial is the way to give 4 balls to 2 people so that every of them has at least 1 ball.