Number of garlands which could be formed by using $3$ flowers of same type and $12$ flowers of other type. Question : Find number of garlands which could be formed by using $3$ flowers of same type and $12$ flowers of other type.

By garland I mean a perfectly circular one .

All $3$ flowers are alike of one type, all other $12$ are alike of different type.

An attempt at the solution.

Firstly we arrange all the $3$ of one type flowers then arrange the other ones, now after arranging the $3$ my circle got divided into $3$ parts and the sum of the flowers in that part has to be $12$.
Let the three parts be $a, b$, and $c$, so
$$a+b+c=12$$
Now the number of integral solutions (without considering the order as flowers are alike) of this equation will be my answer it should be $14\choose 2$.
However, it is wrong. Somebody help me out.
 A: Your approach is correct if the three parts are linear or their order mattered like giving flowers to $3$ people. But here because it is a circle, the total order is not important here. out of the $12$ (say black) flowers, having $3,6,3$ flowers in each part is the same as $3,3,6$ and $6,3,3$ flowers in each part. so you have to remove these repetitions. 
To do this, you should consider all cases of having different numbers of black flowers in each part and count all the repetitions only once. 
A: The numbers that you are after are called compositions (with 0s allowed, rotated such that the largest number comes first, and are not of the form a + b + a with a and b distinct). The compositions of 12 into 3 parts are:
12 + 0 + 0
11 + 1 + 0
11 + 0 + 1
10 + 2 + 0
10 + 1 + 1
10 + 0 + 2
9 + 3 + 0
9 + 2 + 1
9 + 1 + 2
9 + 0 + 3
8 + 4 + 0
8 + 3 + 1
8 + 2 + 2
8 + 1 + 3
8 + 0 + 4
7 + 5 + 0
7 + 4 + 1
7 + 3 + 2
7 + 2 + 3
7 + 1 + 4
7 + 0 + 5
6 + 6 + 0
6 + 5 + 1
6 + 4 + 2
6 + 3 + 3
6 + 2 + 4
6 + 1 + 5
5 + 5 + 2
5 + 4 + 3
5 + 3 + 4
4 + 4 + 4  
So there are 31 of these. 
