How many routes around town? I was always rubbish at maths, so I could really use some help here.  
THE PROBLEM:
Imagine a town with 12 buildings, one of which is home.  I need to figure out how many ways you could visit anywhere from 1 to all 12 buildings in any order.  So at the very least, you are visiting 2 buildings (say, home and building 1).  You could also visit home, building 3, building 11, building 7 and then building 9.  
Is this a combinatorics problem?  A permutations problem?  If so, what do these frightening words mean?  Thanks for helping out the maths-challenged. 
 A: You start at home. You now have $11$ different buildings to go to. On the next turn, you have $10$ choices, because you cannot visit home or the first building you chose. This pattern continues. 
So if you choose to visit $n\leq 12$ buildings (not counting home), you have 
$$11\times 10\times \dots \times (12-n)$$
possible routes. This can be written as $(11)_n$, which denotes the falling factorial. To get your answer, simply sum over all $1\leq n< 12$. Making the generalization $11\rightarrow x$, it turns out that this sum can be written as (thanks Mathematica)
$$\sum_{n=1}^{x}(x)_n=e\Gamma(x+1,1)-1,$$
where $\Gamma(\cdot,\cdot)$ is the incomplete Gamma function.
For $x=11$, this results in $108505111$ possibilities all in all.
A: Assuming that you are always starting from home, you can visit any of the $11$ other buildings in any order / number (ranging from $0$ other buildings to all $11$ of them). 
For some fixed number of $k$ buildings that we wish to visit, there are $\binom{11}{k}$ ways to pick which $k$ we visit, and $k!$ ways to order them. Summing this over all possible $k$ gives us our answer:
$$W = \sum_{k=0}^{11} \binom{11}{k}k! = \sum_{k=1}^{11} \frac{11!}{(11-k)!} = 108505112$$
If the order isn't important then you'd just remove the $k!$ part and the answer would be $2^{11}$ via the binomial theorem.
A: Skip to the end for the answer, or review all of this to help your understanding of the mathematics.
Explaining the language (background maths to help)
A combination is a collection of events which may occur in any order.
A permutation is an ordered collection of events.
Imagine the set of events S={a,b,c}
How many sub-sets consisting of two events are there? {a,b} and {a,c} and {b,c}.
Ok, but what if we require the order to play a part? Then we would have


*

*{a,b}

*{b,a}

*{a,c}

*{c,a}

*{b,c}

*{c,b}


Understanding the mathematics of the problem
To choose K of the N buildings, we use the notation $\binom{N}{K}$.  If we were to select 2 of 4 possible events (where order doesn't matter ) we would write $\binom{4}{2}$.  The actual calculation is performed by $\binom{N}{K}=\frac{N!}{K!(N-K)!}$. The exclamation mark ! denotes factorials.  Take a positive integer like 6. $6!=\prod^6_{i=1}i =6\times 5\times 4\times 3\times 2 \times 1$.  You just multiply all positive integers from the starting number to 1, inclusive. 


*

*$4!=1\times 2 \times 3 \times 4=24.$

*$3!=3 \times 2 \times 1=6$


An important part to note is that $0!=1$ 
SO... $\binom{4}{2}=\frac{4!}{2!(4-2)!}=6$.  There are 6 combinations when selecting any 2 elements from our set of 6 elements, where we do not require order.
What if we DO require order...? Well, if we have K elements, there are $K!$ permutations.
To find the number of permutations for K of N events, we would find $\binom{N}{K}\times K!=\frac{N!}{K!(N-K)!}\times K! = \frac{N!}{(N-K)!}$ The example i gave with events a,b,c and how many permutations (order) when selecting two of the elements, I listed the 6 permutations.  You can use the formula to determine the number of permutations instead of listing them.
Understanding the problem

Imagine a town with 12 buildings, one of which is home. I need to figure out how many ways you could visit anywhere from 1 to all 12 buildings in any order. So at the very least, you are visiting 2 buildings (say, home and building 1). You could also visit home, building 3, building 11, building 7 and then building 9.

Note the phrase:

how many ways you could visit anywhere from 1 to all 12 buildings in any order

Chris has since specified that the order of visiting the buildings IS important.*
It's asking about permutations!
$$\sum^{12}_{K=1}\binom{12}{K}K!$$
to find the number of permutations where you select anywhere from one to all of the buildings to visit.  Since you start at building 1, you are interested in 11 of the buildings.
$$\sum^{11}_{K=1}\binom{11}{K}K!=108505112$$
