Error in attempting the combinatorics problem

Question

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.

Answer 1

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

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

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

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

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

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

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

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

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

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

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

$(4,4),(0,0,0),(1)$

Answer 2

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-$(4,4),(0,0,0),(1)$

Answer 3

$a+ 2b + 3c = 15, a\ge 1, b\ge 1, c\ge 1$

$(10,1,1)\\ (8,2,1)\\ (7,1,2)\\ (6,3,1)\\ (5,2,2)\\ (4,1,3)\\ (4,4,1)\\ (3,3,2)\\ (2,2,3)\\ (1,1,4)\\ (2,5,1)\\ (1,4,2)$