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5

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 $n$.


4

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 $[a,b)$?


2

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 ...


1

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.


1

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.


1

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


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!!.


1

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.


1

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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