Put $7$ balls into $7$ cells. Probability that exactly $2$ cells containing $3$ balls?

Question

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?

Answer 1

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.