How many different (circular) garlands can be made using $3$ white flowers and $6m$ red flowers? This is my first question here. 
I'm given $3$ white flowers and $6m$ red flowers, for some $m \in \mathbb{N}$. I want to make a circular garland using all of the flowers. Two garlands are considered the same if they can be rotated to be the same. My question is: how many different garlands can I make?
I tried an approach in which I divided the question into three parts:


*

*No. of ways in which the white flowers are all together

*No. of ways in which two white flowers are together

*No. of ways in which no two white flowers are together


Can you please help?
The answer is $3m^2+3m+1$ in the book that I saw
Thanks

I first placed all the white flowers creating 3 spaces, each between a pair. The question can now be seen as the number of non-negative integer solutions of
$x+y+z=6m$ 
the number of solutions can be given as
$\binom{6m+3-1}{3-1} = 18m^2 + 9m + 1$ 
but this $18m^2 + 9m +1$ contains three types of solutions: 
$αβγ,ααβ,ααα$ where 

*

*$αβγ$ is counted 6 times $(3!=6)$ 

*$ααβ$ is counted thrice $(\frac{3!}{2!} = 3)$ 

*$ααα$ is counted once.
But, since it is a garland, we need solutions of each form counted only once.
Let $N(αβγ$) represent the number of solutions of the type $αβγ$.
Therefore, the required answer is 
$N(αβγ)+N(ααβ)+N(ααα)$ 
where $N(ααα)=1$(obvious)
For the solutions of the type $ααβ$ we can find the number of solutions of the equation $2x+y=6m$.
The number of solutions of this equation will consist of solutions of the type $ααβ$ and $ααα$.
The number of solutions will be = $\frac{6m}{2} +1 = 3m+1$ (Number of solutions for 2x+y=N)
$=>N(ααβ)+N(ααα)=3m+1$ 
Now,
$18m^2+9m+1=1+3(3m)+6N(αβγ)$ 
Hence, $N(αβγ)=3m^2$ 
Therefore, the required expression, $N(αβγ)+N(ααβ)+N(ααα)=$

$3m^2+3m+1$

