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How many ways you can order the numbers $0, 1, 2, 3,..., 12$ using each number exactly once, such that the sum of two adjacent numbers are not greater than $13$? (This is a first round's question of the four rounds of Bangladesh Mathematical Olympiad for class $11$–$12$.)

For example, these are some orderings which satisfy the condition:

  • $0, 12, 1, 11, 2, 10, 3, 9, 4, 8, 5, 7, 6$
  • $12, 1, 11, 2, 10, 3, 9, 4, 8, 5, 7, 6, 0$
  • $1, 12, 0, 11, 2, 10, 3, 9, 4, 8, 5, 6, 7$
  • $11, 2, 10, 1, 12, 0, 3, 9, 4, 8, 5, 6, 7$
  • $6, 7, 2, 11, 0, 12, 1, 10, 3, 5, 8, 4, 9$
Sayan Dutta
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Aleph-null
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1 Answers1

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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 $P(12)=6P(10)=36P(8)= ... =6^5.$