Problem:
If $n$ balls are distributed at random into $r$ boxes (where $r \geq 3$), what is the probability that box $1$ at exactly $j$ balls for $0 \leq j \leq n$ and box $2$ contains exactly $k$ balls for $0 \leq k \leq n$ ?
Answer:
Let $p$ be the probability that we seek. First we consider a special case. If $j + k > n$ then $p = 0$. Let $p_1$ be the probability that
a ball is placed in box $1$. Let $p_2$ be the probability that a ball is placed in box $2$. Let $p_3$ be the probability that a ball is placed in box other than box $1$ and box $2$.
\begin{align*}
p_1 &= \frac{1}{r} \\
p_2 &= \frac{1}{r} \\
p_3 &= \frac{r-2}{r}
\end{align*}
Now we have a multinomial distribution.
\begin{align*}
P &= \left( \frac{n!}{j!k!(n-j-k)!} \right) \left( \frac{1}{r}\right) ^j \left( \frac{1}{r}\right) ^k \left( \frac{r-2}{r} \right)^{n - j - k} \\
P &= \left( \frac{n!}{j!k!(n-j-k)!} \right) \left( \frac{1}{r}\right) ^{j+k} \left( \frac{r-2}{r} \right)^{n - j - k} \\
P &= \left( \frac{n!}{j!k!(n-j-k)!} \right) \frac{ (r-2)^{n-j-k} } {r^n}
\end{align*}
So for example, if we have: $j = 2$, $k = 2$, $n = 8$ and $r = 8$ then
\begin{align*}
P &= \left( \frac{8!}{2!2!(8-2-2)!} \right) \frac{ (8-2)^{8-2-2} } {8^8} = \left( \frac{8!}{4(8-2-2)!} \right) \frac{ (6)^{4} } {8^8} \\
P &= \left( \frac{8!}{4(4)!} \right) \frac{ (6)^{4} } {8^8} = \left( \frac{8(7)(6)(5)}{4} \right) \frac{ 2^4(3^4) } {8^8} \\
P &= \frac{ 8(7)(6)(5)( 2^4)(3^4) } { 4(8^8) } = \frac{ 2(7)(6)(5)( 2^4)(3^4) } { 4(8^7) } \\
P &= \frac{ 7(6)(5)( 2^4)(3^4) } { 2(8^7) } = \frac{ 7(6)(5)( 2^3 )(3^4) } { (8^7) }\\
P &= \frac{ 7(6)(5)(3^4) } { 8^6 } = \frac{ 17010 } { 262144 } \\
P &= \frac{8505 } { 131072 } \\
P &\doteq 0.064888
\end{align*}
I think I have it right now. Do I?
You have corrected the mistakes that were your initial solution. Your work is now correct.