Astrological Sign Doors There are 12 doors, each labeled with an astrological sign. There are 12 people each representing one of the signs (No one has the same sign). How many ways are there to line up all 12 people in such a way that no person is standing in front of their correct sign?
How I was thinking of this, you have 11 choices for person 'A', but for person 'B' they would have either 11 choices, if person 'A' was in front of the door with their sign, or 10 if person 'A' wasn't. This repeats with person 'C' having 10 or 9 choices, down to the last person having only 1 choice. But I am unsure how to represent this in an actual equation. Is this the correct way to be thinking about this problem?
 A: As noted by saulspatz, you are looking for the number of derangements of a $12$-element set. Your thinking looks correct, but breaking things down by cases gets tedious. 
A formula can be found with inclusion-exclusion. Note that your situation can be reformulated as finding the number of permutations of $\{1,2,3,4,5,6,7,8,9,10, 11, 12\}$ where no element is mapped to itself. 
In the more general case, we are working with the set $[n]=\{1, 2, ..., n\}$. Let $S_n$ be the set of permutations of $[n]$, and let $A_i$ denote the set of all permutations of $[n]$ where $i\in [n]$ is mapped to itself. The complement form of the principle of inclusion-exclusion tells us 

$$\Big|S_n\setminus\bigcup_{i=1}^nA_i\Big|=\sum_{I \subset [n]}(-1)^{|I|}\Big|\bigcap_{i\in I}A_i\Big|.$$

Note that $S_n\setminus\bigcup_{i=1}^nA_i$ is the set of permutations of $[n]$ where no elements are mapped to themselves (i.e. the set of derangements).
So we have 
\begin{align}\Big|S_n\setminus\bigcup_{i=1}^nA_i\Big|
&=\sum_{I \subset [n]}(-1)^{|I|}\Big|\bigcap_{i\in I}A_i\Big|\\
&=\sum_{I \subset [n]}(-1)^{|I|}(\text{# permutations where each element of $I$ is mapped to itself})\\
&=\sum_{I \subset [n]}(-1)^{|I|}(n-|I|)! \\
&=\sum_{k =0}^n\sum_{\substack{I \subset [n];\\ |I|=k}}(-1)^{|I|}(n-|I|)!\\
&=\sum_{k =0}^n\sum_{\substack{I \subset [n];\\ |I|=k}}(-1)^{k}(n-k)!\\
&=\sum_{k =0}^n (-1)^{k}(n-k)!\sum_{\substack{I \subset [n];\\ |I|=k}}1\\
&=\sum_{k =0}^n (-1)^{k}(n-k)!(\text{# of k-element subsets of $[n]$})\\
&=\sum_{k =0}^n (-1)^{k}(n-k)!\binom{n}{k}\\
&=\sum_{k =0}^n (-1)^{k}(n-k)!\frac{n!}{k!(n-k)!}\\
&=n!\sum_{k =0}^n(-1)^k\frac{1}{k!}.
\end{align}
In your case, when $n=12$, the number of derangements is $$12!\sum_{k =0}^{12}(-1)^k\frac{1}{k!}.$$
