$1^{st|}$ element of $X$ can map to any $1$ element of $Y$ i.e. $n$ choices.
$2^{nd}$ element of $X$ can map to any $1$ element of remaining $n-1$ elements of $Y$ i.e. $n-1$ choices.
$3^{rd}$ element of $X$ can map to any $1$ element of remaining $n-2$ elements of $Y$ i.e. $n-2$ choices.
.......
$(n-1)^{th}$ last element of $X$ can map to any $1$ element of remaining $2$ elements of $Y$ i.e. $2$ choices.
$n^{th}$ element of $X$ can map to only last remaining element of $Y$ i.e. $1$ choice.
So number of bijective function possible $=n*(n-1)*(n-2)......3*2*1 = n!$
Hence Option $D.$ is the correct answer.