How many different (circular) garlands can be made using $3$ white flowers and $6m$ red flowers?

My answer would be $\frac{1}{3}\left(\binom{6m+2}{2}-1\right)+1$.

$\binom{6m+2}{2}$ is the number of ways of writing $6m$ as the sum of three non-negative integers.

We count the one case where all the values are the same seperately. That yields one garland.

The other cases, the equations:

$$6m=a+b+c=b+c+a=c+a+b$$ are all the same garland, so we have to divide those cases by $3$.

Simplified:

$$\begin{align} \frac{1}{3}\left(\binom{6m+2}{2}-1\right)+1& = \frac{1}{3}\left((3m+1)(6m+1)-1\right)+1\\ &=6m^2+3m+1 \end{align}$$

in general, if there were $p$ white flowers and $pn$ red, with $p$ prime, then the number of garlands would be:

$$\frac{1}{p}\left(\binom{np+p-1}{p-1}-1\right)+1$$

In this case, $n=2m$ and $p=3$.

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.

First I will dramatically overcount. Then I will overcompensate for my overcounting. Then I will compensate for my overcompensation to reach the final answer.

Imagine the garland as a fixed circle, with a total of $3+6m$ positions for flowers. Then the number of possible garlands is simply the number of ways to choose the $3$ positions of the white flowers. This is the binomial coeffient $$\binom{3+6m}{3} = \frac{(6m+1)(6m+2)(6m+3)}{6}$$

But we've agreed that garlands that can be rotated to be the same fixed garland are considered the same garland. Each garland can be rotated in $3+6m$ ways, so we have to divide our answer above by $3+6m$, to get

$$\frac{(6m+1)(6m+2)}{6}$$

This is N.F.Taussig's answer, but it's missing a little something. This is the one garland that remains the same when rotated by an angle less than $2\pi$. This special garland has the white flowers arranged in an equilateral triangle, and any rotations by $2\pi/3$ do not change the garland. There are $1+2m$ fixed versions of this garland, and we've counted them each $1/(3+6m)$ times. That means we've counted $\frac{1+2m}{3+6m} = 1/3$ of these garlands, when there is actually a full $1$ garland of this type. We need to add $2/3$ to our previous answer, to get our final answer of

$$\frac{(6m+1)(6m+2)}{6}+\frac{2}{3} \,= \,\boxed{6m^2 + 3m+1} \,\,\text{ garlands}$$