Number of surjections from A to B if |B|=|A|-1 In trying to figure out the above problem I looked at Number of surjections from $\{1,...,m\}$ to $\{1,...,n\}$
and i'm fairly confused on a few things. 
Following op's argument, if we look at {1...m} to {1...n} then there are 
$n^m$ total functions. This makes sense to me. But these include non subjective functions where 1,2,3... elements in the co-domain have no mapping. So we subtract non subjective function:
$n \choose 1$$(n-1)^m$ -> set $S_1$ of functions missing 1 element
$n \choose 2$$(n-2)^m$ -> set $S_2$ of functions missing 2 elements. 
But then op makes the comment" but how many times did we count this in the previous count?" This I don't get. These functions are ones where 2 elements in the co-domain have no input that maps to them. The previous ones consisted of functions where only 1 element in the co-domain had no input that mapped to them. So how could there be duplicates? Any function you choose from $S_1$ will differ from another in $S_2$ as functions in $S_2$ will have an additional element in co-domain with no mapping.
Also is there a way to do the given problem 

Find the number of surjections from A to B if $|B|=|A|-1$

without the inclusion-exclusion thing or if not could someone explain it.
 A: 
But then op makes the comment" but how many times did we count this in the previous count?" This I don't get. These functions are ones where 2 elements in the co-domain have no input that maps to them. The previous ones consisted of functions where only 1 element in the co-domain had no input that mapped to them. So how could there be duplicates?

Not quite. Remember that $n^m$ counts all functions from $\{1,\dots,m\} \to \{1,\dots,n\}$, not just the surjective ones. So the first number: $\binom n 1 (n - 1)^m$ counts functions:
$$ f : \{1,\dots,m\} \to \{1,\dots,n\} \setminus \{i\}$$
for all $n$ choices of $i$. This includes non-surjective functions. Or in other words: we guarantee that $i$ isn't in the range but not that the range is exactly $\{1,\dots,n\} \setminus \{i\}$.

Without inclusion/exclusion, think of it as follows. You're trying to count objects that look like this:
$$ \begin{array}{c|c|c|c|c|c|c|c}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline
5 & 3 & 2 & 1,4 & 6 & 8 & 7 & 9
\end{array} $$
where $f(i) = j$ if $i$ appears below $j$.
Instead, what you should think of is this:
$$ \begin{array}{c|c|c|c|c|c|c|c}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline
5 & 3 & 2 & 1 & 6 & 8 & 7 & 9 \\
  &   &   & 4 &   &   &   &
\end{array} $$
There are two ways to decide which element is on top or on bottom, so we will divide by $2$ later to compensate.
How many ways are there to choose which elements go in those rows? And once those elements are chosen, how many ways are there to choose each row?
The answer is: there are $n$ choices for the elements (choose which element of $\{1,\dots,n\}$ goes in the bottom row). For the top row, you have a surjection from an $n - 1$ element set to an $n - 1$ element set so there are $(n - 1)!$ choices here. For the bottom row, there are $n - 1$ spots where you can put that element. So in total, we have
$$ \frac{n \cdot (n - 1)! \cdot (n - 1)}{2} = \frac{n!(n-1)}2 $$
surjections from $\{1,\dots,n\} \to \{1,\dots,n-1\}$.
