Find the probability function of $N$ and $X_1$ Scatter $3$ balls in $3$ cells, let $N$ be the number of taken cell, let $X_1$ be the number of balls in the first cell

Find the probability function of $N$ and $X_1$

My start:
\begin{array}{|c|c|}
\hline &1&2&3\\\hline
  0 & &&0 \\\hline
  1 &  \\\hline
 2 & &&0 \\\hline
 3& &0&0 \\\hline
\end{array}
I'm stuck here
 A: Throwing the balls in the cells is the same as assigning a number to each ball. For the first ball you have $3$ choices, for the second ball $3$ and for the third $3$. Example (111) means all three balls go to cell 1 etc. So each outcome is equally likely and therefore, since there are $27$ outcomes, the probability of each is equal to $1/27$. Now,
\begin{align}P(X_1=0, N=1)&=P((222)\text{ or } (333))=\frac{2}{27}\\P(X_1=0, N=2)&=P((322),(232),(223) \text{ or } (332),(323),(233))=\frac{6}{27}\\P(X_1=1,N=1)&=0
\\P(X_1=1, N=2)&=P((122),(212),(221) \text{ or } (331),(313),(133))=\frac{6}{27}
\\P(X_1=1, N=3)&=P((123),(132),(213),(231),(312),(321))=\frac{6}{27}
\\P(X_1=2, N=1)&=0
\\P(X_1=2, N=2)&=P((112),(121),(211) \text{ or } (113),(131),(311))=\frac{6}{27}
\\P(X_1=3, N=1)&=P((111))=\frac1{27}
\end{align}
This is not the most efficient way, but is secure (you avoid binomials by treating (223) different as (232) but the cost is that you must count outcomes carefully). 

Check that it is correct (not criminally wrong): All the probabilities add up to $1$.
