Number of onto mappings from set {1,2,3,4,5} to the set {a,b,c} This is based on the question at here
I want to know how many onto functions are there from set $A={1,2,3,4,5}$ to set $B={a,b,c}$.    
This is how I did it:
First i tried to find the functions which are not onto
case 1:
All in A maps to a single element of B
There are 3 ways for this  
case2:
When mappings are made for only two elements in range.
First we have to select 2 out of 3 elements.That can be done in $3\choose 2$$=3$ways.
Consider mappings to only (a,b).
Number of ways when only one item maps to a=5
Number of ways when only two items maps to a=$5\choose 2$
Number of ways when only three items maps to a=$5\choose 3$   
Number of ways when only four items maps to a=$5\choose 4$
Total mappings only to (a,b)=30.
Hence in case 2 total non onto mappings are$=30*3=90$  
By case1 and case2 Total non onto mappings are $90+3=93$  
Therefore onto mappings are $3^5=93=150$   
My question is in the given answer mappings for case2 is obtained as , there are 2^5 = 32 possible functions, so we have 3*32 = 96 functions here that aren't onto.    
What's wrong with my method?
Can someone please tell me where I have done wrong?
 A: $2^5$ also counts the maps with only one element in the image. When you then multiply by $3$ you double count the functions that send only to one element. 
For example the function $f_a$ (that sends everything to $a$). Is counted in the case $(a,b)$ and $(a,c)$. Since there are $3$ such functions we have to subtract $3$ to the outcome and we get $96-3=93$ which you had before.  
A: I'll consider $\{1,2,3\}$ instead of $\{a,b,c\}$ (it's easier for the notation)
For $i\in\{1,2,3\}$, let $A_i$ denote the set of all function $f:\{1,...,5\}\to\{1,2,3\}\backslash \{i\}$. A function $f:\{1,...,5\}\to\{1,2,3\}$ is onto if it doesn't belong to $\bigcup_{i=1}^3 A_i$. By inclusion exclusion formula,
$$\left|\bigcup_{i=1}^3 A_i\right|=\sum_{\emptyset\neq I\subset \{1,2,3\}}(-1)^{|I|-1}\left|\bigcap_{i\in I}A_i\right|$$
For any non-emptyset, $I\subset \{1,2,3\}$, $$\left|\bigcap_{i\in I}A_i\right|=(3-|I|)^5,$$ because $\bigcap_{i\in I}A_i$ is the set of all function $f:\{1,...,5\}\to\{1,2,3\}\backslash I$. Therefore,
$$\left|\bigcup_{i=1}^3 A_i\right|=\sum_{\emptyset\neq I\subset \{1,2,3\}}(-1)^{|I|-1}(3-|I|)^5=\sum_{i=1}^3(-1)^{i-1}\binom{3}{i}(3-i)^5.$$
We conclude that the number of onto function is
$$3^5-\left|\bigcup_{i=1}^3 A_i\right|=\sum_{i=0}^3(-1)^i\binom{3}{i}(3-i)^5=...$$
