Probability: Balls in baskets I'm self learning and I stumbled upon the following exercise but I'm not sure if I solved it correct as I'm very new to this.
Problem: 7 balls fall independently into 7 baskets. Let $X_i$ = number of baskets containing $i$ balls. What is the distribution of $X_3$?
Solution:
$A$ = $\{$ Basket contains 3 balls $\}$
$B_j$ = $\{$ A ball falls into basket $j$ $\};j=1,2...7$
$P(B_j) = \frac17$
$P(A) = \binom73 P(B_j)^3P(\overline B_j)^4 = 35\cdot \frac17^3(1-\frac17)^4 = \frac{5\cdot 6^4}{7^6}$
Is this correct so far? I was wondering whether I should multiply the above by 7 since $j$ can be positioned in 7 different combinations but I concluded that since the event $B_j$ doesn't specify that there are other baskets (can be understood as if a ball either goes into basket $j$ or not at all) so I didn't do it. Is that correct?
Now the distribution:
$P(X_3 = 0) = \binom 70 P(A)^0P(\overline A)^7$
$P(X_3 = 1) = \binom 71 P(A)^1P(\overline A)^6$
$P(X_3 = 2) = \binom 72 P(A)^2P(\overline A)^5$
$P(X_3 = 3) = \binom 73 P(A)^3P(\overline A)^4 =N/A$ since $3\cdot 3 > 7$
$P(X_3 = 4) = \binom 74 P(A)^4P(\overline A)^3 =N/A$ since $4\cdot 3 > 7$
$P(X_3 = 5) = \binom 75 P(A)^5P(\overline A)^2 =N/A$ since $5\cdot 3 > 7$
$P(X_3 = 6) = \binom 76 P(A)^6P(\overline A)^1 =N/A$ since $6\cdot 3 > 7$
$P(X_3 = 7) = \binom 77 P(A)^7P(\overline A)^0 =N/A$ since $7\cdot 3 > 7$
Is this correct? 
What is the expected way to prove that from $X_3$ to $X_7$ it's N/A if this was on an exam? Do I even have to write these cases into the table?

As true blue anil stated, the counting method is to count ways to sort balls into a pattern times the permutations of that pattern.
I suggest using multinomial coefficients† to ensure your counting logic matches up.   For instance: the event $X_3=2$ can be satisfied by any pattern of 3 balls in each of two boxes and one ball in another.   First we count ways to assign balls to boxes in that pattern, then ways to arrange the boxes; all divided by $7^7$.
$$\begin{align}\mathsf P(X_3=2) =&~ 7^{-7}{\tbinom{7}{3,3,1\color{silver}{,0,0,0,0}}\tbinom{7}{2,1,4}}
\\[1ex] =&~ 7^{-7}\tbinom{7}{3}\tbinom{4}{3}\tbinom{7}{2}\tbinom{5}{1}
\\[1ex] =&~ \tfrac{6!^2}{7^5~3!^2~2!}
\end{align}$$
Note: the additional zeroes are redundant, and included only to demonstrate that the pattern contains four empty boxes.
Similarly we build the probability for $X_3=1$:
$$\begin{align} \mathsf P(X_3=1) =&~ 7^{-7}\left(\tbinom{7}{3,1,1,1,1\color{silver}{,0,0}}\tbinom{7}{1,4,2}+\tbinom{7}{3,2,1,1\color{silver}{,0,0,0}}\tbinom{7}{1,1,2,3}+\tbinom{7}{3,2,2\color{silver}{,0,0,0,0}}\tbinom{7}{1,2,4}+\tbinom{7}{3,4\color{silver}{,0,0,0,0,0}}\tbinom{7}{1,1,5}\right)
\\[1ex] =&~ \tfrac{6!^2}{3!~7^5}\left(\tfrac{1}{4!~2!}+\tfrac{1}{3!~2!^2}+\tfrac{1}{2!^3~4!}+\tfrac{1}{4!~5!}\right)
\end{align}$$
We could do the same for $X_3=0$, however there are many more possible patterns and it is easier to just note that it is the complement of the union of the above two events.
$$\begin{align}\mathsf P(X_3 = 0) =&~ 1-\mathsf P(X_3=1)-\mathsf P(X_3=2)
\end{align}$$ 

† The multinomial coefficient is: $\dbinom{x+y+z+\ldots}{x,y,z,\ldots} = \dfrac{(x+y+z+\ldots)!}{x!~y!~z!~\cdots} = \dbinom{x+y+z+\ldots}{x}\dbinom{y+z+\ldots}{y}\dbinom{z+\ldots}{z}\cdots$
