If A and B are among 20 persons who sit at ra | vedclass

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(C) Total number of ways to arrange $20$ persons around a round table is $(20-1)! = 19!$. Fix person $A$ at one position. There are $19$ remaining sea

Vedclass · Accepted Answer

$\frac{2}{19}$

Vedclass · Answer

(C) Total number of ways to arrange $20$ persons around a round table is $(20-1)! = 19!$. Fix person $A$ at one position. There are $19$ remaining seats for person $B$. For there to be exactly $6$ persons between $A$ and $B$,$B$ must sit in a specific position relative to $A$. Counting $6$ seats clockwise from $A$,the $7$th seat is occupied by $B$. Counting $6$ seats counter-clockwise from $A$,the $7$th seat is also occupied by $B$. Thus,there are $2$ favorable positions for $B$ out of $19$ possible seats. Therefore,the probability is $\frac{2}{19}$.

If $A$ and $B$ are among $20$ persons who sit at random along a round table,then the probability that there are exactly six persons between $A$ and $B$ is

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