(B) The number of ways to select $5$ persons out of $8$ is ${}^8C_5$.Now,the number of ways to arrange $5$ guests at a round table is $(5-1)! = 4!$.Th

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(B) The number of ways to select $5$ persons out of $8$ is ${}^8C_5$.Now,the number of ways to arrange $5$ guests at a round table is $(5-1)! = 4!$.The number of ways to arrange the remaining $3$ guests at another round table is $(3-1)! = 2!$.Thus,the total number of ways is ${}^8C_5 \times 4! \times 2!$.Calculation: ${}^8C_5 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.Total ways $= 56 \times 24 \times 2 = 56 \times 48 = 2688$.Hence,option $(B)$ is correct.

Class 11 Mathematics Permutation and Combination Circular permutations

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$A$ person invites $8$ guests to a dinner and places $5$ of them at one table and the remaining $3$ at another,both the tables being round. The number of ways in which the guests can be arranged is

AP EAMCET 2019 All AP EAMCET

A
$40320$
B
$2688$
C
$8064$
D
$13440$

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$A$ person invites $8$ guests to a dinner and places $5$ of them at one table and the remaining $3$ at another,both the tables being round. The number of ways in which the guests can be arranged is

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