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Say the number of elements in each of the inner sets is $q.$ If I understand correctly we are working with the unlabeled species $$\mathfrak{M}_{=nm/q}(\mathfrak{M}_{=q}(\mathcal{Q}_1+ \mathcal{Q}_2+\mathcal{Q}_3+\mathcal{Q}_4+\cdots+\mathcal{Q}_n)).$$ It follows using the Polya Enumeration Theorem that the desired value is given by $$[Q_1^m Q_2^m ... 0 These are the Stirling coefficients of the second kind. If the sum of elements in sets must be equal the problem is a lot harder. 2 If you're asking for a formula for the function, I'm afraid there is no straightforward answer. Sloane's OEIS tells us that A065205(n) = A033630(n) - 1. That leads me to$$p_\sigma(n) = f(n, n, 1) - 1$$where$$f(n, m, k) = f(n, m, k + 1) + f(n, m - k, k + 1)0^{n \bmod k}$$if k \leq m, else 0^m. (By the way, I think p(n) is too plain and ... 0 Note: The formulation of the problem seems to be incorrect. Nevertheless here's a hint for a 6-sided die. The generating function for tossing one die once is x^1+x^2+\ldots+x^6, the exponent of x indicating the result of the throw. Let's denote with A(x) the generating function for tossing r dice. We observe, applying the formula for finite ... 4 Number of elements in X_i is n(X_i). We know that$$n(X) = n(x_1) + n(x_2) + n(x_3) = 150n(x_1) = 4 n(x_2)$$and$$ n(x_3) = 5 n(x_2).$$This system of equations can be solved by substituting n(x_1) and n(x_3) to the first equation. 4 This is exactly the content of http://oeis.org/A065205 (as Will Jagy kindly points out, the values of p(n) for n \leq 1000 are listed in http://oeis.org/A065205/b065205.txt); for the first abundant numbers n, p(n) (I'll suppress the suggested subscript) is given by:$$\begin{array}{cc} n & p(n)\\ \hline 12 & 2 \\ 18 & 2 \\ 20 & 1 \\ ...

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When we break a given permutation in $S_n$ into a product of disjoint cycles we obtain a partition of $n$. As you have defined, let $P_k$ be the number of cycles of length $k$ in such a cycle decomposition. Define the type of a permutation to be the vector $P = (P_1, P_2, ..., P_n)$. Note the bijection between types and partitions. Let's count $C_P$, the ...

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hint: look at http://lipn.univ-paris13.fr/~duchamp/Books&more/Macdonald/%5BI._G._Macdonald%5D_Symmetric_Functions_and_Hall_Pol%28BookFi.org%29.pdf pg 24 (2.14) giving the $z_\lambda$ ; your expression is $z_\lambda/n!$ known as the inverses of the class sizes of the symmetric group $S_n$. Example: for $S_5$ we get class sizes 24, 30, 20, 20, 15, 10, 1 ...

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Actually in your particular example you can remove the "all of them between [1;50]" requirement since any partition of $60$ into $5$ distinct positive parts cannot have a part greater than $50$ (the other four must be at least $1+2+3+4$). In your particular example, the answer is $2611$. When Java applets worked on the internet, I would have pointed you at ...

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This is quite hand-wavy but I think that's acceptable considering you want intuition. There is $1$ key gap in the result, but maybe someone can come along and fill that in nicely. First the notation: $N(n)$ is the number of complete compositions of $n$ $N(n,a)$ is the number of complete compositions of $n$ with $a$ as the largest term $N(n,a,i)$ is the ...

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Let's say someone has worked out all the solutions. Let the solutions queue at two doors. If a solution has an $S_m$, they go through the left door, otherwise they go through the right door. The solutions that go through the left door all have an $S_m$ - at least one. The solutions that go through the right door have no $S_m$. So the doors have split the ...

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For any $n > 0$, let $p(n)$ be corresponding number of partitions. $q(n)$ be the corresponding sum $\displaystyle\;\sum_m \prod_k \frac{1}{\lambda_{k,m}!}$. Recall $p(n)$ is the number of solutions for $(x_1, x_2, x_3 \ldots ) \in \mathbb{N}^{\mathbb{Z}_{+}}$ of the equation: $$x_1 \cdot 1 + x_2 \cdot 2 + x_3 \cdot 3 + \ldots = n$$ and its ...

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In general, for sums over structures rather than simple integer ranges you have a few options: Use notation indicating the structure. Here, for example, you could follow George Andrews in using $\lambda \vdash n$ to denote that $\lambda$ is a partition of $n$, and write $$\sum_{\lambda\, \vdash \, n}f(\lambda_1, \ldots, \lambda_n, n)$$ NB I've used ...

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