Feller's Problem II.10.9

Hi, group,

Semi-retiree here with newfound interest in probability, slowly working my way through the problems in Wm Feller's excellent book. Problem 9 in section II.10 is: If n balls are placed at random into n cells, find the probability that exactly one cell remains empty.

Here's my original solution:

(i) There are n^n possible distributions of n balls into n cells so the probability of each distribution is n^(–n).

Now we count the number of distributions that have exactly one empty cell.

(ii) Select one of the n balls to leave out-- there are n ways to do this;

(iii) then there are n! ways to distribute the remaining (n–1) balls into the n cells such that exactly one cell remains empty;

(iv) finally we can place the leftover ball into any of the (n–1) cells that already contain one ball;

(v) therefore, multiplying (ii), (iii) and (iv), there are n*n!*(n–1) distributions of n balls in n cells with exactly one empty cell;

(vi) multiplying by (i), the required probability is n*(n–1)*n!*n^(–n).

My answer is exactly TWICE the answer Feller gives at the end of his book:
C( n,2 )*n!*n^(–n).

Where am I wrong, and what is my mistake? I've Roman-numbered the steps in my reasoning above so you can refer to each step easily.

Please note that I'm NOT asking for the correct solution. I can get Feller's result as follows: From the n balls, there are C( n,2 ) ways to choose two with which to form a "dumbbell," thereby reducing the problem to placing (n–1) items-- the remaining (n–2) balls plus the dumbbell-- into n cells, and that equals n!. I want to know where my original logic failed, since many of my solutions are often off by a factor of two, which I eventually figure out, but I can't seem to figure this one out.