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

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Question

(b) The total number of ways in which $n$ persons can sit at a round table $ = (n - 1)\,\,!.$
$	herefore $ Favourable number of cases $ = 2\,\,!\,\,

Vedclass · Accepted Answer

$(n - 3)\,\,:\,\,2$

Vedclass · Answer

(b) The total number of ways in which $n$ persons can sit at a round table $ = (n - 1)\,\,!.$
$	herefore $ Favourable number of cases $ = 2\,\,!\,\,(n - 2)\,\,!$
Thus the required probability $ = \frac{{2\,\,!\,\,(n - 2)\,\,!}}{{(n - 1)\,\,!}} = \frac{2}{{n - 1}}$
Hence the odds against are $(1 - p):p$ or $(n - 3):2.$

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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