How many different ways can you distribute $5$ apples and $8$ oranges among six children if every child must receive at least one piece of fruit? How many different ways can you distribute $5$ apples and $8$ oranges among six children if every child must receive at least one piece of fruit? If there is a way to solve this using Pólya-Redfield that would be great, but I cannot figure out the group elements.
I know this is a duplicate of: How many different ways can you distribute 5 apples and 8 oranges among six children?. But I can not comment on this or contact the member who explained the task. 
Could someone explain in more detail how to apply this, especially how to evaluate the sums? 
Maybe someone has more examples?  Or even a book with solved exercises? 
The main Problem is i dont know what i need to know in order to apply it.
Application 1: How many distinct circular necklace patterns are possible with n beads, these being available in k colors. 
It could be solved by Burnside's lemma but I also have no clue to to apply it. 
 A: I don't think this is a case for Polya theory. In the following I propose a generating function approach. I'm assuming the apples indistinguishable, the oranges indistinguishable, but the children distinguishable.
Assume for the moment that there is an unlimited supply of apples and oranges. An individual child can receive $j$ apples and $k$ oranges, whereby $j+k\geq1$. This option will enter with weight  $x^jy^k$ into the following calculation. The formal sum of all options for one child then is
$$f(x,y):={1\over1-x}{1\over1-y}-1={x+y-xy\over(1-x)(1-y)}\ .$$
In the product $f(x,y)\cdot f(x,y)$ each option $x^{j'}y^{k'}$ of a first child is multiplied with each option $x^{j''}y^{k''}$ of a second child and stored as $x^{j'+j''}y^{k'+k''}$. Collecting terms  of same combined degrees $(j,k)$ then counts the number of allocations where the two children together have received $j$ apples and $k$ oranges. It follows that the combined options for all six children  sum up to $$f^6(x,y)=(x+y-xy)^6\>\sum_{j=0}^\infty{-6\choose j}(-x)^j\>\sum_{k=0}^\infty{-6\choose k}(-y)^k\ ,$$
and the number $N$ we are after is the coefficient of $x^5y^8$ in this expansion. We therefore can restrict the first sum to $j\leq5$ and the second sum to $k\leq8$ and then let Mathematica do the work. The answer is $N=70\,608$.
A: Here is another "elementary" answer, somewhat similar to the answer of @user84413.
Let $j\in[5]$ be the number of children receiving at least one apple. Given $j$, we can do the following:


*

*Choose $j$ from the six children in ${6\choose j}$ ways, and give each of them an apple.

*Distribute the remaining $5-j$ apples arbitrarily 
among the $j$ chosen children. According to "stars and bars" (S&B) this can be done in 
$${j+(5-j)-1\choose j-1}={4\choose j-1}$$
ways.

*Give an orange to each of the $6-j$ children that has not received an apple.

*Distribute the remaining $8-(6-j)=2+j$ oranges arbitrarily among the six children. According to S&B this can be done in
$${6+(2+j)-1\choose 6-1}={7+j\choose5}$$
ways.
The number $N$ of admissible allocations of the fruits is therefore given by
$$N=\sum_{j=1}^5{6\choose j}\cdot{4\choose j-1}\cdot{7+j\choose5}=70\,608\ .$$
Given the small numbers occurring here the last computation can be done  with pencil and paper, as in user84413's answer.
A: If we consider the distributions of the 5 apples, we have 7 possibilities:
$a) \; 5 \; \;b) \;4+1  \;\;c)\; 3+2 \;\; d)\; 3+1+1 \;\; e) \;2+2+1 \;\; f) \;2+1+1+1 \;\; g) \;1+1+1+1+1$
For each of these cases, we have the following number of ways to distribute the apples and oranges
(multiplying the number of ways to distribute the apples by the number of ways to distribute the remaining oranges after each child without an apple has been given an orange):
$a)\;\;6\cdot\binom{8}{5}=336$
$b)\;\;6(5)\cdot\binom{9}{5}=3780$
$c)\;\;6(5)\cdot\binom{9}{5}=3780$
$d)\;\;6\binom{5}{2}\cdot\binom{10}{5}=15120$
$e) \;\;6\binom{5}{2}\cdot\binom{10}{5}=15120$
$f)\;\;6\binom{5}{3}\cdot\binom{11}{5}=27720$
$g)\;\;6\cdot\binom{12}{5}=4752$
This gives a total of $\color{blue}{70,608}$
