There are 5 gentlemen and 4 ladies to dine at | vedclass

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Question

(B) To ensure no $2$ ladies sit together,we first arrange the $5$ gentlemen around the circular table. The number of ways to arrange $n$ distinct obje

Vedclass · Accepted Answer

$2880$

Vedclass · Answer

(B) To ensure no $2$ ladies sit together,we first arrange the $5$ gentlemen around the circular table. The number of ways to arrange $n$ distinct objects in a circle is $(n-1)!$.Thus,$5$ gentlemen can be seated in $(5-1)! = 4! = 24$ ways.After seating the gentlemen,there are $5$ gaps created between them (as shown by the $X$ marks in the figure).Since no $2$ ladies should be together,we must place the $4$ ladies in these $5$ available gaps.The number of ways to choose and arrange $4$ ladies in $5$ gaps is given by the permutation formula $^nP_r = \frac{n!}{(n-r)!}$.Here,$n=5$ and $r=4$,so the number of ways is $^5P_4 = \frac{5!}{(5-4)!} = \frac{120}{1} = 120$.Therefore,the total number of ways is $24 	imes 120 = 2880$.

There are $5$ gentlemen and $4$ ladies to dine at a round table. In how many ways can they seat themselves so that no $2$ ladies are together?

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