Find the probability that exactly two cells remain empty, one is occupied by three balls and the rest contain each one ball. • $n$ numbered balls are placed at random into n distinguishable cells. Find
the probability that exactly two cells remain empty, one is occupied by
three balls and the rest contain each one ball.
Please can anyone give me a detailed explanation to this.
I said that the first $n-3$ balls have $nPn-3$ ways of being put into boxes without filling a box more than $1$ ball , that leaves us with $3$ boxes empty and $3$ balls not used, then I multiply by $3$ to choose which one of the $3$ boxes will be filled by the $3$ balls. By that, I should get the wrong answer of $$\frac{3 \cdot nP(n-3)}{n^n}$$ 
 A: To count the number of ways, you can look at it as rolling a $n$ sided die $n$ times, and find the product of two multinomial coefficents for choose $\times$ place, of the pattern $3-0-0-1-1-...$
viz. $\binom{n}{1,2,(n-3)}\binom{n}{3,0,0,1,1,1...} = n\binom{n-1}{2}\binom{n}{3}(n-3)!$
and Pr $= \dfrac{n\binom{n-1}{2}\binom{n}{3}(n-3)!}{n^n}$
Added another way:


*

*Choose boxes having 3 balls, and no balls in $\binom{n}1\binom{n-1}2$ ways.
[The rest automatically become boxes for singles]

*Choose balls for the "three" box in $\binom{n}3$ ways

*Permute the singles in their boxes in $(n-3)!$ ways

*Pr $= \dfrac{n\binom{n-1}{2}\binom{n}{3}(n-3)!}{n^n}$, as before.
A: Okay so after long discussion with the closest professor I could find , it turns out that ONE of my many answer was correct lol .  Anyway to those who would like to hear the answer .  First we want to choose $2$ special boxes that will be empty out of $n$ boxes, $\binom{n}{2}$. Then we want to choose $3$ special balls that will form $1$ big ball, $\binom{n}{3}$. Then we would like to permute the $n-2$ balls in $n-2$ boxes, $(n-2)!$ . And the total number of ways we could have put the balls without constraints was $n^n$.
Thus the answer would be 
$$\frac{\binom{n}{3} \cdot \binom{n}{2} \cdot (n-2)!}{n^n}$$
A: I know this is a late answer, but i just saw this when i was studying an old exam and thought I might share the answer given in the answer key:

*

*choose 2 out of n cells (these cells are empty) :nC2

*choose from the leftover cells 1 cell (it will have 3 balls): (n-2)C1

*choose 3 balls from the n balls (to put in cell#2): nC3

*the leftover cells must have each 1 ball, so (n-3)P(n-3) or simply (n-3)!

*divide by n^n (total number of ways of filling n cells with n balls.
After some simplifying using the general fromula of nCk; we get:

$$\frac{n!n(n-1)(n-2)}{12n^n}$$
A: Your hypothesis gives you exactly how the balls are distributed. All that remains is the way the boxes are arranged. You can count all the possible ways the two empty boxes and the box containing  $3$ balls can "move around" among the $n$ boxes. 
Then divide this by all possible combinations, which is $n+1$ choose $n-1$, if I am not mistaken. 
