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4

The easy way to prove this seems by example. The easiest way to make sure that no pair gets seated wrongly, is to make sure that there is no pair at all. You have five classes, just seat one of each like so: ABCDE Then, keep seating such clusters untill you are done: For 2x5 BCD A E E A DCB For 4x5 BCD A E E A D B C ...


3

I'll only show how to obtain a linear recurrent relation from the problem, it has been explained at various places how to solve these. Let: $a(n)$ be solutions (ways how to change $n$) using only $1$c, $b(n)$ solutions using only $1,2$c and at least one $2$c, $c(n)$ solutions using all three coins and at least one $3$c. Clearly each solution fits ...


3

What you have to find is just: ...


2

Let $y_1=x_1-1, y_2=x_2-2,\text{ and } y_3=x_3-3$. Then $y_1+y_2+y_3=14$ $\;\;$ where $y_1\le3, y_2\le8, y_3\le9 \text{ and } y_i\ge0$ for $1\le i\le3$. Let S be the set of solutions without the upper bounds on the $y_i$, and let $A_1$ be the set of solutions with $y_1\ge4$, $A_2$ be the set of solutions with $y_2\ge9$, and $\;\;\;\;A_3$ be the set of ...


2

The following method could be explained to any student that understands equations for lines and knows how to sum an arithmetic sequence. However, you may find it at least as tedious as the method you already have! The Idea Think of points $(x,y)$ in the plane as representing "$x$ 2-cent coins and $y$ 3-cent coins".   Now, the grid points in the first ...


2

Given that there are 5 types of person, each of whom sits to the left (or right) of ONE other person, this relationship can be shown as a square (person by neighbour). Also, the seating is going to be a continuous sequence where the 'left' person of the current pair will be the 'right' person of the next pair. Each neighbour relationship can be numbered 1 ...


2

A To expectation of the minimum used label, $K$, we first measure the probability that all the balls being among the top $n-k$ boxes. That is, that the minimum label will be greater than some value $k$. In the total space each of $n$ balls has a choice of $n$ boxes ($n^n$). In the restricted space each of $n$ balls has a choice of $n-k$ boxes $(n-k)^n$. ...


2

Permutations with repetition of n elements are permuations where the first element is repeated a times, the second b times, the third c times, ...: $$\frac{n!}{a! b! c! \dots}$$ We have in this case $6$ elements where the first element is repeated $3$ times and the second also $3$ times. So there are $$\frac{6!}{3! 3!}$$ possible numbers.


2

The teams are $A, B,.. G, H$. The match-ups are denoted by the the lines/connections (edges) drawn between the teams (vertices). The four hideous colours - Green, Blue, Yellow and Purple - represent the four sports. Should be easy to interpret. There are no multi-edges and hence no teams play each other twice. Each vertex has exactly four distinct edges ...


1

The question here is in how many different ways you can select the $n$ positions of the ones from the $2^m$ available positions in the vector. That's called combinations.


1

The reason why this is so is because of the statement in the previous paragraph that "if the equality holds then each pair $(n, \ell_j-n)$ has a gap and a non-gap" (I'm renaming the variable here to prevent confusion). Since we have that $k = 2j-r\in L$, it must be the case that $\ell_j-k\in H$. Meanwhile, the reason for this statement is a basic ...


1

I am getting: $\left(x+x^2+x^3+x^4+...+x^{15}\right)\left(x^{15}+x^{20}+x^{25}+...+x^{85}\right)$ for the first kid. $(\left(x^3+x^4+x^5+...+x^{15}\right) \left(x^{15}+x^{20}+x^{25}+...+x^{45}\right))^3$ for the next 3 kids. Then multiply both of them and check the coefficient of $x^{100}$


1

It is possible to shorten a bit the argument by noticing that $$\operatorname{Res}\left(\frac{1}{(1-z)(1-z^2)(1-z^3)},z=1\right)=-\frac{17}{72},$$ hence the contribute given by the simple poles $-1,\omega,\omega^2$ lying on the unit circle cannot exceed $\frac{17}{72}$, and is exactly equal to $\frac{17}{72}$ when $6\mid n$. The pole in $z=1$ is a triple ...



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