From $6$ different novels and $3$ different dictionaries, $4$ novels and $1$ dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangements is

The correct option is B at least $1000$
Let $N_1{N}_2{N}_3\dots N_6$ be novels and $D_1{D}_2{D}_3$ be dictionaries
The number of ways ${=}^6{C}_4{\times }^3{C}_1\times (4!)$

$NNDNN$
Here, ${}^6{C}_4$ denotes the selection of $4$ novels out of $6{,}^3{C}_1$ denotes the selection of $1$ dictionary out of $3$ and $(4!)$ denotes the rearrangement of $4$ novels as dictionary will always be in the middle.
$=15\times 3\times 24=1080$
Hence, option 'D' is correct.