Number of surjections from $\{1,2,3,4,5,6\}$ to $\{a,b,c,d,e\}$ Where $A = \{1,2,3,4,5,6\}$ and $B = \{a,b,c,d,e\}$.
My book says it's:


*

*Select a two-element subset of $A$.

*Assign images without repetition to the two-element subset and the four
remaining individual elements of $A$.


This shows that the total number of surjections from $A$ to $B$ is $C(6, 2)5! = 1800$.
I'm confused at why it's multiplied by $5!$ and not by $4!$. Also in part 2, when we assign images, do they mean images in $B$? 
 A: How many ways can $A$ be partitioned into $5$ blocks?
Answer: $\binom{6}{2} = 15$
Given any $5\text{-block}$ partition of $A$, in how many ways can the blocks be bijectively assigned to the $5$ element set $B$?
Answer: $5! =120$
How many surjective functions from $A$ onto $B$ are there?
Answer: $15 \times 120 = 1800$
A: Think of it this way:
There is a pair of terms that get mapped to the same element.  Call that pair $\alpha $.  There are four terms remaining.  Call them $\beta,\gamma,\delta$ and $\epsilon $.
There are ${6\choose 2} $ possible pairs that can be $\alpha $.  
And we must map $\alpha,\beta,\gamma,\delta,\epsilon $ to $a,b,c,d,e $.  There is $5! $ ways to do that.
A: (i). Select  a $2$-member  $A_1\subset A.$ There are $\binom {6}{2}$ ways to do this. Select a $1$-member  $B_1\subset B.$ There are $\binom {5}{1}$ ways to do this. There are $\binom {6}{2}\binom {5}{1}$ such pairs $(A_1,B_1)$ and for each pair  there is a set $F(A_1,B_1)$ of $4!$ surjections $f:A\to B$ such that $\{f(x):x\in A_1\}=B_1.$
And if $(A_1, B_1)\ne (A'_1, B'_1)$ then the sets $F(A_1,B_1), F(A'_1,B'_1)$ are disjoint.
So there are at least $\binom {6}{2}\binom {5}{1}4!=(15)(5)(4!)=(15)(5!)=1800$ surjections. 
(ii). Every surjection $f:A\to B$ belongs to some $F(A_1,B_1)$ so there are at most $1800$ surjections.
In other words: (i) we didn't count any $f$ more than once, and (ii) we didn't fail to count any $f$.
Remark. The "mysterious" $5!$ came from two sources: The product of the number $\binom {5}{1}$  of $B_1$'s and the number $4$!  of bijections from a $4$-member set to another.
