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Suppose $X$ and $Y$ are finite sets, each with cardinality $n$.. The number of bijective functions from $X$ to $Y$ is

- $n^{n}$
- $n\log_{2}n$
- $n^{2}$
- $n!$

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**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.