Error in attempting the combinatorics problem In how many ways can $15$ identical blankets be distributed among $6$
  beggars such that everyone gets at least one blanket and two
  particular beggars get equal blankets and another three particular
  beggars get equal blankets.


Attempt:

First, $6$ blankets are given and then $9$ blankets are left. 
After that, according to me there are only two possibilities: 


Possibility $1$: 


First set of particular beggars receives: $(3,3)$

Second set of particular beggars receives: $(1,1,1)$ 


Possibility $2$:


First set of particular beggars receives: $(0,0)$

Second set of  particular beggars receives: $(3,3,3)$ 

However answer given is $12$.

Please tell me my mistake.
$(0,0),(3,3,3),(0)$

$(0,0),(2,2,2),(3)$

$(0,0),(1,1,1),(6)$  

$(0,0),(0,0,0),(9)$ 

$(1,1),(2,2,2),(1)$ 

$(1,1),(1,1,1),(4)$ 

$(1,1),(0,0,0),(7)$ 

$(2,2),(0,0,0),(5)$

$(2,2),(1,1,1),(2)$ 

$(3,3),(0,0,0),(3)$  

$(3,3),(1,1,1),(0)$ 

$(4,4),(0,0,0),(1)$
For set of 2 and 3 beggars of equal number, and remaining one:  

1-$(0,0),(0,0,0),(9)$
2-$(0,0),(1,1,1),(6)$
3-$(0,0),(2,2,2),(3)$
4-$(0,0),(3,3,3),(0)$
5-$(1,1),(0,0,0),(7)$
6-$(1,1),(2,2,2),(1)$
7-$(1,1),(1,1,1),(4)$
8-$(2,2),(0,0,0),(5)$
9-$(2,2),(1,1,1),(2)$
10-$(3,3),(0,0,0),(3)$
11-$(3,3),(1,1,1),(0)$
12-$(4,4),(0,0,0),(1)$