Question
Suppose $n(\geq 3)$ persons are sitting in a row. Two of them are sclected at random. The probability that they are not together is(a) $1-\frac{2}{n}$(b) $\frac{2}{n-1}$(c) $1-\frac{1}{n}$(d) $\frac{2}{n}$
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This can be done using the combination formula, which is given by $nC2 = \frac{n(n-1)}{2}$. Show more…
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