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A lift has $N$ stops ($1,2,3,4,...,N$), hence have $N(N-1)$ distinct rides of travelling from floor $A$ to floor $B$ such that $A\neq B$. How many arrangements of these rides form a continuous trip that starts from floor 1 and ends at floor 1?

I was lost trying to solve it. Anyone has an idea?

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  • $\begingroup$ isnt it just rearanging all the $n-1$ upper floors? so $(n-1)!$, no? $\endgroup$
    – Shaq
    Commented Feb 2, 2021 at 21:51
  • $\begingroup$ Welcome to MSE! Please review the Meta Read and enhance your question to provide your motivation/attempts. $\endgroup$
    – Jessie
    Commented Feb 2, 2021 at 21:54

1 Answer 1

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There are $N-2$ possible intermediate stopping points; at floors $2, 3, ..., N-1$.

Each one of these can either be chosen or not and so the total number of arrangements is $2^{N-2}.$

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