If a party of n persons sit at a round table, | vedclass

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(B) The total number of ways in which $n$ persons can sit at a round table is $(n - 1)!$. To find the number of ways where two specified individuals s

Vedclass · Accepted Answer

$(n - 3) : 2$

Vedclass · Answer

(B) The total number of ways in which $n$ persons can sit at a round table is $(n - 1)!$. To find the number of ways where two specified individuals sit together,we treat them as a single unit. This gives us $(n - 1)$ units,which can be arranged in $(n - 2)!$ ways. The two individuals can be arranged among themselves in $2!$ ways. So,the number of favourable ways is $2! 	imes (n - 2)!$. The probability $p$ that they sit together is $\frac{2! 	imes (n - 2)!}{(n - 1)!} = \frac{2}{n - 1}$. The odds against the event are given by $(1 - p) : p$. $(1 - \frac{2}{n - 1}) : \frac{2}{n - 1} = \frac{n - 3}{n - 1} : \frac{2}{n - 1} = (n - 3) : 2$.

If a party of $n$ persons sit at a round table,then the odds against two specified individuals sitting next to each other are

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