Circular permutations with alike objects

Question

There are $6n$ flowers of one kind and $3$ flowers of another kind. How many different garlands can be made using all flowers.

I tried to first arrange the flowers linearly and then joining the endpoints of that linear string. Then I subtracted the cases where there would have been repetition. Through this I am getting the answer as $18n^2+9n+1$ But the answer given is $3n^2+3n+1$

Can someone please provide me with a hint.
Note : A similar question is

Circular permutations with indistinguishable objects

But the answers use something like Polya's Enumeration theorem which I have no idea of.

Answer 1

If the only "symmetry" of the garland is the rotating it, then your answer is right. However if we can obtain one garland from another by flipping it, so there will be less unique garlands.

In the first case note that we have a $1-1$ correspondence between unique garlands and linear alignments of the $6n+3$ flowers which start with a flower of the second kind. To note this just choose one of the flowers of the second kind and consider it as "special". Cut the garland before that flower and open it up to get a linear alignment. Similarly you can glue the end of such a linear alignment to get a unique garland. So as you found out the number of such garlands is:

$$\binom{6n+2}{2} = \frac{(6n+2)(6n+1)}{2} = 18n^2 + 9n + 1$$

In the second case there is a $1-1$ correspondence between unique garlands and linear alignments of the $6n+3$ flowers, starting with a flower of the second kind, but also the sequences of consequtive flowers of the first kind should be in non-deceasing order. In other words you want to find how many partitions of $6n$ are there in $3$ summands. In general this isn't easy to find, but as we have only three summands things get easier.

If all the three summands are same you have only one unique partition.

If two of the summands are same you have $3n$ such paritions. To note this the different term has to be even and it uniquely determines the partition. There are $3n+1$ even numbers between $0$ and $6n$, but we don't count when it's $2n$, as we counted that in the first case.

The case when all summands are different is not so easy to calculate. By Stars and Bars we have $\binom{6n+3}{3}$ ways to write $6n$ as sum of $3$ terms. (This number counts $3+0+3 = 6$ and $3+3+0=6$ as two different sums, but we need to consider them as same). Now this number counts the sums when all numbers are same just once, and when exactly two of them are same it counts them three times, while when all the the numbers are different it counts them $6$ times. So using this the number of partitions with different summands is:

$$\frac 16 \left(\binom{6n+3}{3} - 1 - 3\cdot 3n\right) = 3n^2 $$

So finally the number of partitions of $6n$ in $3$ terms is $3n^2 + 3n+1$, which is the unique number of garlands.

Answer 2

You need to find the number of unordered solutions (a,b,c) to the equation $a+b+c=6n$, where a,b,c are non negative integers.

Case 1: a=b=c=2n, one such tuple

Case 2: exactly two of a,b,c are equal. $3n$ such cases.

Case 3: $a \neq b \neq c$. It'll be $\frac{1}{3!} ({6n+2 \choose 2} -1-{3 \choose 2}3n) = \frac{1}{6}(18n^2)=3n^2$.

So adding all cases, the final answer is $3n^2+3n+1$.