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A permutation of $1,2,3, \ldots, n $ is chosen at random . The probability that $1$ and $2$ are neighbors is -

$(1) \frac {1}{n} $

$(2) \frac {2}{n} $

$(3) \frac {1}{n-1} $

$(4) \frac {1}{n-2} $

I think the correct answer is $(2) $ $ \frac{2}{n}$.

Reasoning : Total number of permutation $=n!$ . Now consider $1,2$ as single object permute $(1,2),3, \ldots, n$. They are $(n-1)!$ in numbers. For each of these permutations $(12)$ and $(21)$ are counted differently, so there are $2(n-1)!$ permutation such that $1,2$ are neighbors . .

So the probability $=\frac {2(n-1)!}{n!}=\frac{2}{n}$

Is my solution correct? If it is not, then supply a proof and possible explanation why my solution is incorrect. Thank you. Your efforts are highly appreciated.

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    $\begingroup$ Correct. _________ $\endgroup$ May 8, 2017 at 15:22

1 Answer 1

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If "$1$ and $2$ are neighbour" means that "$1$ and $2$ are neighboring," the solution you have provided appears to be correct. It is $$\frac{2 \times (n-1)!}{n!}=\frac{2}{n}$$ As you have solved.

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    $\begingroup$ That is exactly what i meant , Thank you ! $\endgroup$ May 8, 2017 at 15:31

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