If n persons are seated around a circular tab | vedclass

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(B) The total number of ways $n$ persons can be seated around a circular table is $(n - 1)!$. The number of ways in which two specific persons sit tog

Vedclass · Accepted Answer

$(n - 3) : 2$

Vedclass · Answer

(B) The total number of ways $n$ persons can be seated around a circular table is $(n - 1)!$. The number of ways in which two specific persons sit together is $2! 	imes (n - 2)!$. The probability $p$ that they sit together is $p = \frac{2! (n - 2)!}{(n - 1)!} = \frac{2}{n - 1}$. The probability $q$ that they do not sit together is $q = 1 - p = 1 - \frac{2}{n - 1} = \frac{n - 3}{n - 1}$. The ratio of unfavorable probability to favorable probability is $q : p = \frac{n - 3}{n - 1} : \frac{2}{n - 1} = (n - 3) : 2$.

If $n$ persons are seated around a circular table,what is the ratio of the unfavorable probability to the favorable probability that two specific persons sit together?

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