Good question.
You cannot apply the same reasoning in the case when we have repetitions, because the $4!$ ways you can arrange $4$ objects assume different objects.
Let's look at your example of choosing $4$ objects from $10$ possible objects (with repetition) and say that our $10$ possible choices are letters A to J.
One possible permutation is ABBA, which is different to the permutation BABA. Now think about how many ways do we have to arrange two As and two Bs. It is not $4!$ (we do not have 4 ways to choose the first letter, nor 3 choices for the second letter etc).
In the extreme case where we have say CCCC (this is one permutation), we also have only one combination.
So when you divide by $r!$, you are potentially overestimating the number of ways you can arrange the objects in a permutation, because you might have repeated objects in this permutation. Moreover, it is difficult to calculate the average arrangements because it depends on the permutation we have each time. So instead of trying to solve the problem this way, we attack it differently (stars and bars for example).
This method is not a problem when repetitions are not allowed because you can easily count the possible arrangements for any permutation. They are always the same, independent of the permutation we have.
You are right, there is definitely a connection between these two counting problems. In fact, the second problem (also known as "balls and urns" or "stars and bar") is really just a variant of the first.
In the first problem, we first treat same letters and distinguishable from one another (e.g. "mississippi" becomes "M$\text{i}_1\text{s}_1\text{s}_2\text{i}_2\text{s}_3\text{s}_4\text{i}_3\text{p}_1\text{p}_2\text{i}_4$"). There are a total of $11$ letters so there are $11!$ total ways to arrange them. However we have clearly overcounted, as the i's, the s's, and the p's are not different from each other. In fact, for a letter that was repeated $x$ times, we have counted every valid arrangement of "mississippi" $x!$ times (do you see why?). Therefore, we must divide out these overcounted arrangements from $11!$, so our final answer is $\frac{11!}{4!4!2!}$.
Suppose we want to put $m$ indistinguishable balls into $n$ distinguishable urns. How many ways are there to do this?
We can line up the $m$ balls in a row from left to right, like this:
$B\text{ } B\text{ } \cdots \text{ }B$
Now imagine that we place $n-1$ partitions (let us denote them with the letter P) in the spaces between the B's to separate them into urns as follows:
- Every ball to the left of the leftmost partition goes into urn 1.
- Every ball between the $i-1$th and $i$th partition goes into urn $i$.
- Every ball to the right of the rightmost partition goes into urn $n$.
Note that it is perfectly fine to have multiple partitions next to each other; this just means that the corresponding urn has $0$ balls.
Because every arrangement is valid, we are in essence rearranging letters of the "word" with $m$ B's and $n-1$ P's. This is exactly the same as the first problem, and our final answer is $\frac{(m+n-1)!}{m!(n-1)!}=\binom{m+n-1}{m}$.
Hope this helped :)
Best Answer
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.