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## New answers tagged partitions

1

Let $\lambda=\langle\lambda_1,\ldots,\lambda_r\rangle$. Let $F$ be the Ferrers diagram for $\mu$. For $k\in[r]$ let $F_k$ be what remains of $F$ when each row longer than $\lambda_k$ has been shortened to length $\lambda_k$. Verify that $\sum_{i,j}\min\{\lambda_i,\mu_j\}=\sum_{k=1}^r|F_k|$. We now determine how many times each column of $F$ is counted in ...

0

We can model the situation with generating functions. In order to do so we consider binary strings consisting of $0s$ and $1s$. Let $$0^\star=\{\varepsilon,0,00,000,\ldots\}$$ denote all strings containing only $0$s with length $\geq0$. The empty string is denoted with $\varepsilon$. The corresponding generating function is ...

1

Suppose otherwise. Write $A = \{a_1, \dotsc, a_n\}$, $B = \{b_1, \dotsc, b_n\}$, and $C = \{c_1, \dotsc, c_n\}$, each indexed increasingly, such that $a_1 < b_1 < c_1$. Thus $a_1 = 1$. Set $k = b_1 - 1$, so that $\{1, \dotsc, k\} \subseteq A$. Now, for any $b \in B$, if any of $b+1, \dotsc, b+k$ are in $C$, then we have a forbidden triple. Similarly ...

1

HINT: Number the elements of $S$ from $1$ through $2n$. In order to form a partition of $S$ into $n$ pairs, we can begin by pairing element $1$ with one of the other $2n-1$ elements. That leaves $2n-2$ elements still to be paired.

3

No, there is no known closed form for the partition function. Here is the sequence on OEIS, and also the Wolfram MathWorld page on this subject is very thorough. Although, looking at this MathSE post, I might be wrong ...

-5

we get $$\frac{(1-p^n)^{n+1}}{1-p^n}$$

0

It may interest the reader to see how this can be done using generating functions. Fixing the parameter $k$ we seek to show that $${n\brace k} = \sum_{m=0}^{n-1} {n-1\choose m} {m\brace k-1}.$$ Here we have extended the summation back to zero because the second Stirling number produces zero for those extra values. Recall the species for set partitions ...

0

$S(n,k)$ is the number of partitions of the set $[n]=\{1,\ldots,n\}$ into exactly $k$ non-empty parts. Suppose that $\mathscr{P}$ is such a partition; then there must be a $P\in\mathscr{P}$ that contains the number $n$. The other $k-1$ parts must contain at least $k-1$ and at most $n-1$ elements of $[n]$; why? Let $m=|[n]\setminus P|$, the number of ...

1

$$\sum_{i=0}^{n-1} f(x_i) \, \Delta x_i = \sum_{i=0}^{n-1} \sqrt{\frac{4i^2}{n^2}} \left( \frac{4(i+1)^2}{n^2} - \frac{4i^2}{n^2} \right).$$ Etc. Why don't you know where to begin? Are you unaware of what Riemann sums are? If so, maybe you could ask about that.

0

You can equivalently think of partitions as equivalence relations, with elements in the same piece of the partition labelled as equivalent. So the partition in the first subset $S_1$ is given by the equivalence $$B \sim_1 C,$$ and the partition in the second subset $S_2$ is given by the equivalence $$C \sim_2 D \sim_2 E.$$ We can even suppose that we have ...

0

There are $n^k$ total cells. For a cell to not overlap, we must change all the coordinates and have $n-1$ choices for each one. There are $(n-1)^k$ non-overlapping cells, so $n^k-(n-1)^k-1$ other cells that overlap. For $n=k=3$ that gives $18$

0

Supposing the range of coordinates is from $1$ to $n$, you can count the number of overlapping cells by examining the cell at coordinates $(n, n, \ldots, n)$, that is, $(a_1, a_2, \ldots, a_k)$ where $a_1 = a_2 = \ldots = a_k = n$. Every cell has the same number of overlapping cells, which can be shown for any other cell $C$ by swapping "slices" of the cube ...

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