Distinguishing between combinations and permutations I have a problem with understanding few exercises in combinatorics. I thought a lot about the problems but it seems too confusing:


*

*There are 3 girls in blue dresses, 2 girls in red and 2 girls in white. How many ways are there for making a closed circle of them if the girls in the same color are not distinguishable.

*Mother comes to her 4 kids from another country carrying gifts. She has 5 teddy bears, 3 dinosaurs and 6 trucks, how many ways are there for giving them the toys if there is a possibility that one or all kids don't get anything.

*There is a bakery that sells 3 donuts with strawberry, 4 with chocolate, 6 with vanilla and 2 with creme. How many ways are there to pick 5 donuts and give them to 3 people, each of them must have at least one?
My idea for the first one is to do permutation of group with repeating and divide by the number of times we can rotate the circle so we get rid of same permutations, but I am not too sure.
 A: I'll do 1. using Burnside's Lemma.  The group in question is the cyclic group $C_7$ on $7$ points, and it acts on the $7$ girls by cyclically permuting them.  Burnside's Lemma implies the number of inequivalent arrangements is $$\frac{1}{|C_7|} \sum_{\alpha \in C_7} \text{total nr arrangements fixed by } \alpha$$ or equivalently $$\frac{1}{7} \sum_{\alpha \in C_7} \text{total nr arrangements fixed by } \langle\alpha\rangle.$$
For any non-identity permutation $\alpha \in C_7$, we have $\langle\alpha\rangle=C_7$.  This has the important consequence that, if an arrangement of girls were fixed by the non-identity permutation $\alpha$, all the girls would have to wear the same color dress.  But this is not possible since e.g. some girls have blue dresses and others have red dresses.

We conclude that the number of inequivalent arrangements is $$\frac{1}{7} \text{total nr arrangements fixed by } \mathrm{id}_{C_7}$$ which is $$\frac{1}{7} \binom{7}{3,2,2}=30$$ using the multinomial coefficient.
(Note: this matches G Tony Jacobs's answer, but only because non-identity  $\alpha \in C_7$ stabilize no arrangements.  If the question instead asked for "4 girls in blue dresses, 2 girls in red and 2 girls in blue", G Tony Jacobs's method would give the wrong answer; in fact, it would give a non-integer answer.)
A: *

*For this one, I'd just count the ways to arrange them in a line, and then divide by 7, because all 7 cyclic permutations of the same linear arrangement count as the same circular arrangement. To count linear arrangements, we could just place the blue dresses in $\tbinom 73$ ways, leaving $4$ spots open, then arrange the red dresses in $\tbinom 42$ ways. That give us a final answer of $\frac{1}{7}\tbinom 73 \tbinom 42 = \frac{7!4!}{7(3!4!)(2!2!)} = 30$ arrangements.

*Let's deal with bears first. In a row of $5$ identical bears, there are $6$ spaces to place dividers: before bear $1$, between bears $1$ and $2$, etc. To divide them into four (possibly empty) subsets for the four kids, we need to place $3$ dividers, and it's possible for multiple dividers to occupy the same space. There are $6^3$ ways to do this. Following this reasoning, our formula for the total in this question should be:
$(5+1)^3(3+1)^3(6+1)^3 = 168^3 = 314,432$ If we're also allowing that some (or all) toys might not go to any of the four children, then we can increase the exponent from $3$ to $4$.

*I can't think of a good way to do this one without splitting it up in to a ton of cases. The fact that the number of each donut type is limited, with some of them quite small, makes it tricky. The number of ways to pick $5$ donuts doesn't seem too hard, but the number of ways to distribute them will depend on which donuts were picked.
