<s>[INST] Put $7$ balls into $7$ cells. Probability that exactly $2$ cells containing $3$ balls? 
1.46. Seven balls are distributed randomly into seven cells. What is the probability that the number of cells containing exactly $3$ balls is $2$?

I am getting different answer from this solution manual. My argument is the following:
In order for us to have an arrangement where there are exactly $2$ cells with $3$ balls, we can follow the following procedure. We first decide which $3$ balls we want to put them together among all $7$ balls, and then decide to which cell we want to put them in. This gives us $\binom{7}{3} \binom{7}{1}$. Now, among the left $4$ balls, we choose $3$ balls to put them together, and choose one cell among the remaining $6$ empty cells. This gives us $\binom{4}{3} \binom{6}{1}$. Finally, we are left with one ball, and we have $5$ choices regarding where to put it. Putting them together, we have
$$P(X_3 = 2) = \frac{\binom{7}{3} \binom{7}{1} \binom{4}{3} \binom{6}{1} 5}{7^7}$$
But the solution manual says the answer should be
$$\frac{\binom{7}{2}\binom{7}{3}\binom{4}{3}5}{7^7}$$
Who is wrong and why?
 [/INST] The manual is right.  The difference between your solution and the books solution is a factor of $2.$ That is ${7\choose1}{6\choose1} = 2{7\choose 2}$
You have placed 3 balls in one of 7 cells and 3 balls in one of 6 remaining.  But you have double counted at this step.
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