Suppose that n people are seated in a single row of n theate

Solution provided by shrouded1 is correct.

Suppose that n people are seated in a single row of n theater seats in a random manner. What is the probability that two particular people will be seated next to each other?

Consider n people: \(x_1\), \(x_2\), ..., \(x_n\).

Probability=(# of favorable outcomes)/(total # of outcomes)

# of arrangement of these n people in a row would be \(n!\) and this would be total # of outcomes.

Now, suppose we are asked to find the probability that \(x_1\) and \(x_2\) will be seated next to each other. Consider them as one unit, glue them like: {\(x_1,x_2\)}, {\(x_3\)}, {\(x_4\)}..., {\(x_n\)} - in this case they will be next to each other. We have \(n-1\) object now and # of arrangement of these objects would be \((n-1)!\), but within this unit \(x_1\) and \(x_2\) can be arranged in 2! # of ways: either {\(x_1\), \(x_2\)} or {\(x_2\), \(x_1\)}. So total # of favorable outcomes would be \(2!*(n-1)!\).

\(P=\frac{2!*(n-1)!}{n!}=\frac{2}{n}\).

Hope it's clear.

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