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 always try to promote the Polya Enumeration Theorem because it has many applications from simple to very sophisticated, and with this in mind I would like to apply it here. Once you have placed the three white flowers there are three slots for the $6m$ red ones, with every slot holding some number of red flowers including the case of zero red flowers. The group acting on these slots is the cyclic group on three elements, whose cycle index can be calculated with pen and paper and turns out to be
$$Z(C_3) = \frac{1}{3}( a_1^3 + 2a_3).$$
It follows the answer is given by
$$[z^{6m}] Z(C_3)\left(\frac{1}{1-z}\right).$$
This is
$$\frac{1}{3} [z^{6m}] \frac{1}{(1-z)^3}
+ \frac{2}{3} [z^{6m}] \frac{1}{1-z^3}$$
which is
$$\frac{1}{3} {6m+2\choose 2} + \frac{2}{3} 
= \frac{1}{6} (6m+2)(6m+1) + \frac{2}{3} 
= 6m^2 + 3m + 1.$$
