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(D) Step $1$: Select $4$ novels out of $6$ novels in ${}^6C_4$ ways.${}^6C_4 = \frac{6 	imes 5}{2 	imes 1} = 15$ ways.Step $2$: Select $1$ poetry bo

Vedclass · Accepted Answer

$1080$

Vedclass · Answer

(D) Step $1$: Select $4$ novels out of $6$ novels in ${}^6C_4$ ways.${}^6C_4 = \frac{6 	imes 5}{2 	imes 1} = 15$ ways.Step $2$: Select $1$ poetry book out of $3$ poetry books in ${}^3C_1$ ways.${}^3C_1 = 3$ ways.Step $3$: Arrange the $4$ selected novels and $1$ poetry book in a row such that the poetry book is in the middle. The arrangement looks like: $(N_1, N_2, P, N_3, N_4)$.The $4$ novels can be arranged in the $4$ available positions in $4!$ ways.$4! = 4 	imes 3 	imes 2 	imes 1 = 24$ ways.Step $4$: Total number of arrangements $= {}^6C_4 	imes {}^3C_1 	imes 4! = 15 	imes 3 	imes 24 = 1080$.

There are $6$ different novels and $3$ different poetry books on a table. If $4$ novels and $1$ poetry book are to be selected and arranged in a row on a shelf such that the poetry book is always in the middle,then the number of such possible arrangements is

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