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The purpose is to find the number of ways to color $2n$ following integers in red and blue, such that if $i$ is red, so is $i-1$.

I tried to use Inclusion-Exclusion principle, but I got stuck in the calculations.

It is a little tricky.

I am trying to count the number of cases where there are $k$ bad tuples $\color{blue}{i-1}\color{red}{i}$.

I need to get ${2n-k\choose k}$ term for this calculation.

Explanations I ran into merely stated that there were $k$ places that are "taken", but it doesn't make much sense as they can't be taken until they are taken, I mean, it's more in retrospect.

How can I explain such a process? Edit: the final answer is very clear to me, but the binomial expression is what I struggle with.

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    $\begingroup$ There are $2n$ ways. If $i$ is red, so is $i-1$, so is $i-2$, $\ldots$. So, you only have one choice, choosing the last red element. $\endgroup$ – Emre May 16 '16 at 14:18
  • $\begingroup$ I don't look for the final answer. I am to present an identity. That's why I need help with the binomial expression. $\endgroup$ – Meitar May 16 '16 at 14:23
  • $\begingroup$ Are you sure you posted the question correctly, I don't see why you use blue $i-1$ next to red $i$ where your question suggest something totally different. $\endgroup$ – Emre May 16 '16 at 14:26
  • $\begingroup$ Because there is an identity I am to arrive at, which uses inclusion and exclusion. Looking at forbidden coloring, I can use the principle. How come this principle not relevant? $\endgroup$ – Meitar May 16 '16 at 14:27
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    $\begingroup$ The problem as stated simply has nothing whatever to do with inclusion-exclusion! Which is why it seems likely you've stated the problem wrong... $\endgroup$ – David C. Ullrich May 16 '16 at 14:31
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Try it like this you have $k$ BR blocks and $2n-2k$ unit blocks. So, you have $2n-k$ empty spots, and you want to choose $k$ spots out of these $2n-k$ to place BR blocks. B=blue,R=red

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  • $\begingroup$ I am a little confused when it comes to having $2n$ colored blue. It is legal and will leave me with $2n-(k+1)$ R blocks. How can I relate to that? $\endgroup$ – Meitar May 16 '16 at 14:41
  • $\begingroup$ It appears to me that It doesn't matter actually. All channels could be what ever they are... That's the concept of Exclusion and Inclusion.. $\endgroup$ – Meitar May 16 '16 at 14:49
  • $\begingroup$ This is different from your comment. Once you choose the highest number red, everything is determined-blue above because that is the highest red and red below by induction. I don't understand this answer $\endgroup$ – Ross Millikan May 16 '16 at 15:05
  • $\begingroup$ The OP does not want the solution, s/he wants to apply inclusion/exclusion to the problem. $\endgroup$ – Emre May 16 '16 at 15:07
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1 must be blue: If I = 1 then i-1 doesn't exist and therefore isn't red, so it must be blue. If i=2 then I-1 is blue so 2 is blue, and so on. All i Must be blue.

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