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Suppose $S_1,\ldots,S_k$ is such a partition. For each $i$, since $\bigcup S_i \neq A$, there is some element $\alpha_i$ that all sets in $S_i$ miss. Consider now the subset $T = \{\alpha_1,\ldots,\alpha_k\}$, and an arbitrary $U \supseteq T$ of size $k$. None of the families $S_i$ contain $U$, hence $S_1,\ldots,S_k$ is not a partition.

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The number of partitions of a finite set is counted by Bell numbers, summing up Stirling numbers. See a similar question here: Number of equivalence relations on a finite set

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Hint: Consider $$f(x) = \left\{ \begin{array}{cc} \frac{1}{x-b}, & x < b \\ 1, & x \ge b\end{array}\right.$$

2

Hint: since $[b,c]$ is compact and $f$ is continuous, $f$ must be bounded on $[b,c]$; but this doesn't need to be true on $[a,b)$. Can you think of a function that wouldn't be integrable on $[a,b)$?

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Write $$U(P,f)-L(P,f) = \sum (M_i-m_i) (x_{i+1}-x_i) = 0$$ Every term in the sum is nonnegative since both $(M_i-m_i) \geq 0$ and $x_{i+1}-x_i > 0$. The only way the sum evaluates to zero is if... which means $f$ is constant on each interval... finally we make sure endpoints are okay.

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By definition $$m_i=\inf_{x_{i}\le x\le x_{i+1}} f(x),\quad M_i=\sup_{x_{i}\le x\le x_{i+1}} f(x)$$ Since trivially $m_i\le M_i$ and $(x_{i+1}-x_i) >0$, if one $m_i<M_i$ we have that strict inequality holds between $L(f,P)$ and $U(f,P)$, hence $m_i=M_i$ for all $i$. Then as $$m_i\le f(x^*)\le M_i=m_i,\qquad x_i\le x^*\le x_{i+1}$$ so equality holds ...

0

For any partition (n) or (k,1^n-k) alias (k,1,1,..,1) with n-k '1's, the resulting tableaux are always () alias empty tableau or ((1),(2), ...,(n-k)) alias a vertical one-column tableau. Bur for any partition (k,2,1,1,..,1) with n-k-2 '1's, the beheaded tableau can be either ((1,2),(3),..,(n-k-1)) or ((1,3),(2),..,(n-k-1)) depending on T. So, beyond these ...

1

No, it's not. It is only 15*13*...*1, or 15!! We consider one of the teams. It has 15 choices for its partner Then consider another of the remaining 14 teams. It has 13 choices Continue doing this, until we only get 1 choice so the answer is 15*13*...*1=15!!.

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To put it simply, there are $9!$ ways in all, if the same-colored flags are considered distinct. However, they are counted as the same, so we consider how many times is each configuration repeated. There are 3! ways for the white, 2! for the red, 4! for the blue, so the final answer is $\frac{9!}{3!2!4!}$. This is the same as the quantity you gave above. ...

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This link might be helpful: http://stackoverflow.com/questions/3866528/equal-k-subsets-algorithm Unfortunately, k-subset is known to be NP-Hard problem, so no algorithm with polynomial time complexity is known.

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Outline: Let $P=(a,b,c)$ be a partition of $n$ into $3$ parts $1\le a\le b\le c$. Let $\varphi(P)=(n-c,n-b,n-a)$. Show that $\varphi$ gives a one to one correspondence between the partitions of $n$ into $3$ parts and the partitions of $2n$ into $3$ parts with each part less than $n$.

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I am not sure completely but I think this is correct For every partition of 2010 into 10 parts order them in increasing order and to lowest number add $0$ to the next lowest add $1$ and so on. For example: Say the partition is $2+2+2+2+2+2+2+2+2+1992$ it becomes $2+3+4+5+6+7+8+9+10+2001$. Due to this costruction every element in the set of partitions of ...

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Take the Ferrers diagram for a partition of 2010 into 10 parts. Add 9 circles to the first row, 8 to the second row, ..., 1 to the next-to-last row. Now we have a partition of $2010+9+8+\cdots+1=2055$ into 10 parts, and they must all be distinct. This is the bijection you seek, now prove it's one-to-one and onto.

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I am posting a recursive algorithm to print all such combinations. Hope this will be helpful. //prints all ordered combination of 1 and 2 which sum up to n function printAll(n){ //[] is empty list print(n,[]) } function print(n,list){ //base case if(n==0){ output all the elements in list }else{ //passing a new list by appending "1" ...

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A simpler way to state the same is the following: Suppose $A = \{a_0, …, a_{k-1} \}$ is a partition of $n$ into $k$ positive distinct integers. Then take the following partition: $$A' = \{a_0-1, …, a_{k-1}-1\}$$ so $$\sum A' = n-k$$ and $A'$ has either $k$ or $k-1$ distinct positive integers (at most one of them is $0$). This process defines a function ...

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This problem merits special attention because it is an instance of a partition problem that can be solved by the Polya Enumeration Theorem as well as by Ferrer's diagrams. Using the material from this MSE link we have for the sum on the LHS the equality $$\sum_{q=1}^k P_q(n) = \sum_{q=1}^k [z^n] Z(S_q)\left(\frac{z}{1-z}\right) = [z^n] \sum_{q=1}^k ... 2 HINT: Take any partition of n+k into k parts and subtract 1 from each part. What do you get? Is the process reversible? 1 I am not sure, what you refer to as the last example, but there is indeed 1-to-1 correspondence between algebras and partitions at over finite sets. The elements of partitions are atoms of the algebra, that is sets that don't contain any proper non-empty subset which is also an element of algebra. Note that partition$$ E := ...

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