21 friends were invited for a party. Two roun | vedclass

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(D) Step $1$: Select $12$ friends out of $21$ to sit at the first table. The number of ways is $\binom{21}{12}$.Step $2$: The number of ways to arrang

Vedclass · Accepted Answer

$\binom{21}{12} 	imes 11! 	imes 8!$

Vedclass · Answer

(D) Step $1$: Select $12$ friends out of $21$ to sit at the first table. The number of ways is $\binom{21}{12}$.Step $2$: The number of ways to arrange $12$ friends at a round table is $(12-1)! = 11!$.Step $3$: The remaining $9$ friends are to be seated at the second round table. The number of ways to arrange them is $(9-1)! = 8!$.Step $4$: The total number of ways is $\binom{21}{12} 	imes 11! 	imes 8! = \frac{21!}{12!9!} 	imes 11! 	imes 8! = \frac{21!}{12 	imes 9} = \frac{21!}{108}$.

$21$ friends were invited for a party. Two round tables can accommodate $12$ and $9$ friends respectively. The number of ways of the seating arrangements of friends is:

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