<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] The given answer is wrong! The correct one has been found by NCh.
If the question was: find the probability that box 1 contains exactly 3 balls and box 2 contains exactly 2 balls then result is
$$p:=\frac{\binom{10}{3}\cdot\binom{7}{2}\cdot 2^5}{4^{10}}=\frac{315}{2^{12}}$$
$\binom{10}{3}$ ways to choose the balls  to be put in box 1, $\binom{7}{2}$ ways to choose the balls to be put in box 2, and $2^5$ ways to place the $5$ remaining balls in box 3 and 4.
If we multiply the probability $p$ by $4\cdot 3$, the number ordered couples of boxes, we find the given result
$$\frac{945}{2^{10}}.$$
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