<s>[INST] probability that two of these boxes contain exactly $2$ and $3$ balls is 
If $10$ different balls are to be placed in $4$ distinct boxes at random,  then the probability that two of these boxes contain exactly $2$ and $3$ balls is 

What I tried:
Total number of ways $\displaystyle 4^{10}$
probability that two of these boxes contain exactly $2$ and $3$ balls is 
 $\displaystyle \binom{4}{2}\cdot 2^5$
So required probability $$\frac{\binom{4}{2}\cdot 2^5}{4^{10}}$$
But answer given as $\displaystyle \frac{945}{2^{10}}$
 [/INST] We have the following variants to distribute balls into the boxes: 
$$
\underbrace{2+3+0+5}_{A_1}, \quad \underbrace{2+3+1+4}_{A_2}, \quad \underbrace{2+3+2+3}_{A_3}.
$$
And the probability that two of these box contain exactly $2$ and $3$ balls is the sum of probabilities for these variants to distribute balls: 
$$
\mathbb P(A)=\mathbb P(A_1)+\mathbb P(A_2)+\mathbb P(A_3).
$$
Find $\mathbb P(A_1)$. There are $\dfrac{10!}{2!\cdot 3!\cdot 0!\cdot 5!}$ ways to chose balls for each box and $4!$ ways to arrange boxes: box for $2$ balls can be chosen by $4$ ways, box for $3$ balls - by $3$ ways, for $0$ balls - by $2$ ways and the rest box is for $5$ balls. So 
$$
\mathbb P(A_1) = \frac{10! \cdot 4!}{2!\cdot 3! \cdot 0!\cdot 5! \cdot 4^{10}} = \frac{60480}{4^{10}},
$$
The same way 
$$
\mathbb P(A_2) = \frac{10! \cdot 4!}{2!\cdot 3! \cdot 1!\cdot 4! \cdot 4^{10}} = \frac{302400}{4^{10}}.
$$
For $A_3$, we need to chose pair of boxes for two balls by $\binom{4}{2}$ ways instead of $4!$. So 
$$
\mathbb P(A_3) = \frac{10! \cdot \binom{4}{2}}{2!\cdot 3! \cdot 2!\cdot 3! \cdot 4^{10}} = \frac{151200}{4^{10}}.
$$
Finally
$$
\mathbb P(A)=\frac{514080}{4^{10}}=\frac{16065}{2\cdot 4^7}\approx 0,490264893.
$$
</s>