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We take $11$ numbers $0,1,2,\cdots,10$. How many ways are there to arrange those numbers so that sum of two adjacent numbers is not greater than $11$?

I heard these types of problems can be solved using recursion. I would like to see a solution using that.

madness
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1 Answers1

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This is the method I used for another similar problem with numbers $0$ to $12$:

Denote the number by $P(12)$.

There are six different arrangements for the placing of the $12$. It can be at one end and adjacent to $0$ or $1$ or it can be adjacent to both $0$ and $1$ in some order.

Consider a line starting $12,0$. The number of possibilities is then the number of arrangements of $1$ to $11$ with sum not greater than $13$ but (subtracting $1$ from each number) this is the same as $P(10)$.

The line starting $12,1$ is the same since neither $0$ nor $1$ add to greater than $13$ with a remaining number.

In the case of $0,12,1$ occurring in a line, replace this sequence by $1$ and then we again require the number of arrangements of $1$ to $11$ with sum not greater than $13$.

Thus the recursive method is to use $P(n)=6P(n-2)$.