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3

The most direct way to solve this is to consider the complement of the given set - that is, the bijections that do have a fixed points. $$Y=\{f:S\rightarrow S\mid f\text{ is bijective and }f(x)=x\text{ for some } x\in S\}.$$ Notice that if we define, for each $s\in S$ the set $$Y_s=\{f:S\rightarrow S\mid f\text{ if bijective and }f(s)=s\}$$ then we may ...


2

Calculation with small values of $n,d,k\geq 0$ shows the following is valid \begin{align*} \sum_{j=0}^{d}(-1)^{d-j}\binom{d}{k}\binom{n-j-1}{n-d-1}\binom{n-d}{j-k} = \begin{cases} 1&\qquad 0\leq k=d<n\\ 0&\qquad \text{otherwise} \end{cases} \tag{1} \end{align*} Note: It's easy to prove the special case $k=d<n$. In order to show that the ...


2

Yes. It is possible. We need to exploit the fact the difference will always be a multiple of $7$. Insert the card to the first machine 14 times to get $(19,33)$. Put that together with the original card to the third machine, and get $(5,33)$. Repeat the process. Using the first machine go up $(33,61)$, then using the third get $(5,61)$. Using the cards ...


1

The words with three different letters are the rearrangement of "AEP" so are $3!=6$. The words with two different letters are rearrangement of AAP,AAE,EEA,EEP so 12. So i get a total of 18.


1

Substitute $r=d-j$ and use upper negation followed by Vandermonde's Identity: $$\require{cancel}\begin{align}\sum_{j=0}^d(-1)^{d-j}\binom{d}{k} \binom{n-j-1}{n-d-1} \binom{n-d}{j-k} &=\binom dk \sum_{r=0}^d(-1)^r\color{blue}{\binom{n-d-1+r}{n-d-1}}\binom{n-d}{d-k-r}\\ &=\binom dk \sum_{r=0}^d(-1)^r\color{blue}{\binom{n-d-1+r}r}\binom{n-d}{d-k-r}\\ ...


1

Here is another approach. Let $c(n,r)$ be the number of ways of filling an $n\times 1$ block with tiles using precisely $r$ colours. Then $c(n+1,r)=c(n,r)+rc(n,r)+rc(n,r-1)$ where the configuration is extended to the right. The first term corresponds to extending the final block of a configuration of length $n$ so it is one unit longer. The second term ...


1

This is not True in genral ; let's take $S=\{a,b,c,d\}$ and let's take our rows: $\{a \}$ and $\{b \}$ $\{b \}$ and $\{c \}$ $\{c \}$ and $\{d \}$ $\{d \}$ and $\{a \}$ Until here every thing works, this example verifies your assumptions, and because the subsets are all singletons then it verifies the second assumption if and only if we did not ...


1

For the sake of definiteness, let's say you want to play clubs; there are $k$ clubs you don't have above the one you want to play, $l$ clubs you don't have below it, and $m$ trumps you don't have. It's the losing events, not the winning events, that are disjoint. One of two losing events is that your opponent has a higher club, with probability $$ ...



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