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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 was a way to solve this using Pólya-Redfield that would be great, but I cannot figure out the group elements.

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I am too lazy to calculate the numbers of elements with $k$ cycles in $S_8$ but if you do that yourself a solution could work as follows. (I will use this version of Redfield-Polya and $[n]$ shall denote $\{1,\dots,n\}$.)

Let us take $X = [13]$ the set of fruits, where $G= S_5 \times S_8$ acts on $X$ such that the first five apples and the later eight oranges are indistinguishable. Then $$K_n = |[n]^X/G|$$ is the number of ways to distribute these apples and oranges among six distinguishable children. And $$ N_n = K_n -n\cdot K_{n-1}$$ the of ways to distribute these apples and oranges among six distinguishable children such that every child must recieve at least one piece of fruit.

Now by the Theorem $$K_n = \frac{1}{|G|} \sum_{g\in G} n^{c(g)} = \frac{1}{5!\cdot 8!} \left(\sum_{g\in S_5} n^{c(g)}\right)\left(\sum_{g\in S_8} n^{c(g)}\right) = \frac{1}{5!\cdot 8!} \left(\sum_{i\in [2]} d_i n^{i}\right) \left(\sum_{i\in [4]} e_i n^{i}\right),$$ where $c(g)$ is the number of cycles of $g$, $d_i$ the number of permutations of $S_5$ with exactly $i$ cycles and $e_i$ the number of permutations of $S_8$ with exactly $i$ cycles.

The number that we are looking for in the end is $N_6$.

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Wow, just what I was asking; however, I am having a difficult time determining the summations at the end, and does n=6? – poyla fan May 16 '11 at 2:24
I added what you are looking for in the end. If you have further problems with how I manipulated the sums, I might write it down in more clarity later but encourage you to try to figure it out by yourself. If you have problems specifically figuring out $$\sum_{i\in[4]} e_in^i,$$ I hope there might be an easier way to find that but I couldn't come up with it yet. I can give you $$\sum_{i\in[2]}d_in^i = 1 + 84 n + 35 n^2$$ as a reference though as this is a lot easier to calculate. – Peter Patzt May 16 '11 at 10:45

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