Answer $D$
The number of bijective functions are possible $\bf{only\;when}$ the number of elements in both the sets are same.
Its is given that the $X$ and $Y$ are the two sets each with cardinality $n$. $\therefore$ Number of elements are same.
Now, in set $X$
Number of choices for first element = $n$
Number of choices for second element = $n-1$
Number of choices for third element = $n-2$
$\vdots\;\;\;\vdots\;\;\;\vdots$
$\vdots\;\;\;\vdots\;\;\;\vdots$
$\vdots\;\;\;\vdots\;\;\;\vdots$
Number of choices for $(n-1)^{th}$ element = $2$
Number of choices for $n^{th}$ element = $1$
So, total choices are:
$$n(n-1)(n-2)(n-3)\dots\dots3.2.1 = n!$$
$\therefore D$ is the correct option.