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Partition the board into 'odd squares' and 'even squares', like black and white on a chessboard. A coin being on an odd square is equivalent to being on any other odd square, and the the same with even squares. So there are are 4 inequivalent ways the coins can be arranged: OO, EE, EO, OE, where OE means blue on an odd square, black on an even square. For ...

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We treat separately some special cases. a) $1$ and $2$ are each alone. It is clear that there is only one way to do this. Same for the other two pairs, for a total of $3$. b) $1$ is alone and the other two are not, and the other two related cases. We leave this to you, after you have gone through the rest of the solution. c) None of $1$, $2$, $3$ is ...

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Hint: It's the same as the three times the number of ways to partition the set $\{4,5, \dots, 100\}$ into three different parts where as many as two of the parts are allowed to be empty (can you see why?), and a formula for the latter value exists (specifically, you should look into Stirling numbers of the second kind and modify the count to account for the ...

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You can give a bijection argument: Interpret $b(i,j)$ as the number of partitions of $m=i+j$ with greatest part $j$ occuring at least twice. Then the LHS counts the number of partitions of $m$ with greatest part occuring at least twice. Now give a bijection of the set of partitions of $m$ with greatest part occuring at least twice to the set of partitions ...

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James Stirling has the wisdom. A Stirling number of the second kind (or Stirling partition number) is the number of ways to partition a set of $n$ objects into $k$ non-empty subsets. In this case, we multiply by $k!$ since we have labeled subsets. So the number is $$4!\ S(9,4)=186480.$$

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Then partition $P$ will have only four points: $$t_=0,\,\,t_1=1-\delta,\,\,t_2=1+\delta,\,\,t_3=2.$$ Clearly $$U(f,P)-L(f,P)=3\cdot 2\delta-1\cdot 2\delta=4\delta.$$ Hence you need $$0<\delta<2^{1002}.$$

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First problem: We use Inclusion/Exclusion. Call the bags A, B, and C. There are $3^{15}$ ways to place the balls into bags, with no restriction. This is because for every one of the balls, we have $3$ choices for the bag it goes into. However, every bag must contain at least one ball. So we count the number of bad ways to distribute the balls, that is, ...

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Hint Try the following procedure: Select how many balls for each bag, say $k_1 + k_2 + k_3 = n$ where $n$ is your number of balls. How many ways are there to do that? Now how many ways are there to select $k_1$ balls from $n$ balls to go in the first bag? How many ways to pick the $k_2$ balls to go into the second bag from $n - k_1$ balls? How many ways ...

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The difference between a $P_1$ partition and a $P_2$ partition is that a $P_1$ partition cannot contain even numbers congruent to $-2$ (mod $4r$), while a $P_2$ partition cannot repeat odd numbers. So to find a bijection, our first hope would be that these "excluded partitions" somehow map onto each other. The additional fact we have about both $P_1$ and ...

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I think the word partition explain itself. it is something to do with parts. so if P are all the parts of X then definitely when you put all the P together it will form X. SO THE UNION OF ALL P IS EQUAL TO X.

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In order to see the "correct" behavior of the Schur polynomials and to be able to compare different formulas for the Schur polynomials, the number of variables you need is equal to the total number of boxes in the partition (the value $d$, in the Wikipedia notation), not just the number of (nonzero) parts in the partition. Try taking $n=d=4$ in both of your ...

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