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Q.
$18 !$ is not divisible by
Permutations and Combinations
Solution:
$18 !$ has $9$ groups of $2$ consecutive integers
$\Rightarrow 18 !$ is divisible by $(2 !)^{9}$
$18 !$ has $2$ groups of $9$ consecutive integers
$\Rightarrow 18 !$ is divisible by $(9 !)^{2}$
$18 !$ has $3$ groups of $6$ consecutive integers
$\Rightarrow 18 !$ is divisible by $(6 !)^{3},$ hence divisible by $(6 !)^{2}$