"Elegant" way to assign 4 distinct objects into 4 bins

1. Breaking it up a little differently, the number of ways with $k = 0,1,2,3$ of them getting $0$ is $_4C_k \cdot {}_3C_{3-k}$, which gives the sequence $1, 12, 18 = 6 + 12, 4$. Where the difference is that you had the $6$ ways of $2,2,0,0$ and the $12$ ways of $3,1,0,0$ separately. Those are $\frac{4!}{2!2!}$ and $\frac{4!}{1!1!2!}$, respectively.
2. For the arrangements, you are looking at the multinomial coefficients: $\frac{4!}{1!1!1!1!} = 24, \frac{4!}{2!1!1!0!} = 12, \frac{4!}{2!2!0!0!} = 6, \frac{4!}{3!1!0!0!} = 4, \frac{4!}{4!0!0!0!} = 1$
3. Instead of focusing on the people and which objects they get, look at the objects and who they are given to. Each object can be given to any of the four people, without restriction. That means there are $4$ options for the first object, $4$ for the second, and so on, so the total is $4^4$.

Let the objects choose the people . . .

Each object has $4$ choices, and each object's choices are independent of the choices made by the other objects, hence there are $4^4$ ways for the objects to choose the people.

Alternatively, using your cases . . .

• For distribution type $1$, the number of ways is $$4!=24$$ Explanation:
  If we have the people line up to choose, then person $1$ has $4$ choices, person $2$ has $3$ choices, person $3$ has $2$ choices, and person $4$ has $1$ choice.

• For distribution type $2$, the number of ways is $$\binom{4}{1}\binom{4}{2}\binom{3}{2}2!=144$$ Explanation:
  • Choose the person who gets two objects:$\;{\large{\binom{4}{1}}}\;$choices.
  • Choose the two objects for that person:$\;{\large{\binom{4}{2}}}\;$choices.
  • Choose the two people to get the two remaining objects:$\;{\large{\binom{3}{2}}}\;$choices.
  • Distribute the two remainining objects, one to each of the two chosen people:$\;2!\;$choices.

• For distribution type $3$, the number of ways is $$\binom{4}{2}\binom{2}{1}\binom{3}{1}=36$$ Explanation:
  • Choose the two people who get two objects:$\;{\large{\binom{4}{2}}}\;$choices.
  • Choose the person who gets the object labeled #$1$ plus one other object:$\;{\large{\binom{2}{1}}}\;$choices.
  • Choose the other object for that person:$\;{\large{\binom{3}{1}}}\;$choices.

• For distribution type $4$, the number of ways is $$\binom{4}{1}\binom{4}{3}\binom{3}{1}=48$$ Explanation:
  • Choose the person who gets three objects:$\;{\large{\binom{4}{1}}}\;$choices.
  • Choose the three objects for that person:$\;{\large{\binom{4}{3}}}\;$choices.
  • Choose the person to get the one remaining object:$\;{\large{\binom{3}{1}}}\;$choices.

• For distribution type $5$, the number of ways is $$\binom{4}{1}=4$$ Explanation:$\;$Choose the person who gets all four objects:$\;{\large{\binom{4}{1}}}\;$choices.

$\;\;\;$Summing the counts for the cases, the total number of ways is $$24+144+36+48+4=256$$