TITLE: How many different (circular) garlands can be made using $3$ white flowers and $6m$ red flowers?
QUESTION [7 upvotes]: 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

REPLY [4 votes]: I discussed this question with my colleagues and the answer is $3m^2+3m+1$
Writing $6m$  as the sum of three non-negative integers is $\binom{6m+2}{2}$ which is equal to $18m^2+9m+1$
$$6m=a+b+c=b+c+a=c+a+b$$ 
Cases where a,b and c are not different
$0 \le 6m-2i$
We get $i \le 3m$ & $0 \le i$
Number of cases with two terms be same is (3m)*$\frac{3!}{2!}$=9m we count {2m,2m,2m} as 1
Number of cases wirh different value of a,b and c is 
$18m^2+9m+1-[9m+1]$=$18m^2$
When a,b and c there are 3! Case so number of garland forn with different a,b & c is $\frac{18m^2}{3!}=3m^2$
Number of cases with two similar values is 3m
Number of cases with a=b=c is 1.
Hence total number of garland is $3m^2+3m+1$ which is correct and this has been checked by an expert.