A party of 23 persons take their seats at a r | vedclass

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(A) The total number of ways to arrange $23$ persons at a round table is $(23-1)! = 22!$.To find the number of ways two specific persons sit together,

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$10:1$

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(A) The total number of ways to arrange $23$ persons at a round table is $(23-1)! = 22!$.To find the number of ways two specific persons sit together,we treat them as a single unit. Now we have $22$ units to arrange in a circle,which can be done in $(22-1)! = 21!$ ways. The two persons can be arranged among themselves in $2!$ ways.So,the number of favorable ways for them to sit together is $21! 	imes 2!$.The probability $P$ that they sit together is $\frac{21! 	imes 2!}{22!} = \frac{2}{22} = \frac{1}{11}$.The probability that they do not sit together is $1 - \frac{1}{11} = \frac{10}{11}$.The odds against them sitting together are given by the ratio of the probability of not sitting together to the probability of sitting together,which is $\frac{10}{11} : \frac{1}{11} = 10 : 1$.

$A$ party of $23$ persons take their seats at a round table. The odds against two specific persons sitting together are

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