Here's the question (from Conroy's dice problems):
We roll a six-sides die n times. What is the probability that all faces have appeared?
I understand the inclusion-exclusion answer to this question, but I don't understand why my alternate solution doesn't work:
Faces 1-6 can be placed/permuted in ${}_nP_r$ ways with $n=n, r=6$ among the n die rolls. There are $6^{n-6}$ permutations for the remaining die rolls, giving ${}_nP_6 * 6^{n-6}$ different permutations with all faces 1-6 out of a possible $6^n$ permutations. So I get an answer of $\dfrac{{}_nP_6}{6^{6}}$, but apparently this is wrong. What am I missing?
You are over-counting. If a sequence has multiple ways to select six places in the sequence containing the numbers $1,2,\dots,6$, then that sequence is counted multiple times by the formula ${}_{n}P_{6}\cdot 6^{n-6}$.
For example, with $n=8$, consider the sequence $$ 1,1,2,2,3,4,5,6. $$ This is counted exactly four times by your method, because there are four ways to select six entries which are numbered $1,2,\dots,6$.
$\color{blue}1,1,\color{blue}2,2,\color{blue}3,\color{blue}4,\color{blue}5,\color{blue}6$
$\color{blue}1,1,2,\color{blue}2,\color{blue}3,\color{blue}4,\color{blue}5,\color{blue}6$
$1,\color{blue}1,\color{blue}2,2,\color{blue}3,\color{blue}4,\color{blue}5,\color{blue}6$
$1,\color{blue}1,2,\color{blue}2,\color{blue}3,\color{blue}4,\color{blue}5,\color{blue}6$
This shows that ${}_nP_6 \cdot 6^{n-6}$ does not count the sequences you want uniquely, which explains why you are getting the wrong answer.