In how many ways can 4 balls be distributed in 3 distinct boxes when each box may have any number of balls. Also, 2 balls are identical

Question

I know how to do it for $n$ identical ones being distributed in $k$ distinguishable boxes as $\frac{(n+k-1)!}{n!(k-1)!}$ and for $n$ distinguishable ones being distributed among $k$ boxes as $k^n$, but for this case I am confused. I guess the answer should be $\frac{3^2 (2+3-1)!}{2!(3-1)!}$ which comes out to be $\frac{9\times4!}{2!2!}=54$.

Answer 1

Start with 1 ball, one of the two identical ones. It can be distributed into 3 distinct boxes in 3 ways, obviously.

Now take the other identical ball. You can also distribute this one into the 3 distinct boxes in 3 ways. Because the two balls are identical, we note that there are only six possible arrangements. If they weren't identical, there would be nine, by the multiplication rule. But they're identical, so the possible box arrangements are: $\{[2,0,0],[0,2,0],[0,0,2],[1,1,0],[0,1,1],[1,0,1]\}$. Okay. Now we have six ways to arrange the two balls into the boxes.

Now we take the distinct ball. There are, again, three ways to arrange this one into the boxes. By the multiplication rule, we now have $3 \cdot 6 = 18$ ways to arrange the three balls into the boxes.

We have two distinct balls, so we repeat the last step again, for $3 \cdot 18 = 54$ total ways to arrange these four balls.