$\begin{array}{1 1}(A)\;48\\(B)\;52\\(C)\;54\\(D)\;56\end{array} $

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- $nP_r=\large\frac{n!}{(n-r)!}$

Out of 5 digits 4 digit numbers are to be formed

Such numbers are $5P_4$

$5P_4=\large\frac{5!}{(5-4)!}$

$\;\;\;\;\;\;\;=\large\frac{5!}{1!}$

$\;\;\;\;\;\;\;=5\times 4\times 3\times 2$

$\;\;\;\;\;\;\;=120$

When 2 is at unit's place ,then remaining three places are filled in $4P_3$ ways

$4P_3=\large\frac{4!}{(4-3)!}$

$\qquad=4\times 3\times 2$

$\qquad=24$

When 4 is at units place,then 4 digit numbers are again =24

$\therefore$ Even 4 digit numbers =$2\times 24$

$\Rightarrow 48$

Hence (D) is the correct answer.

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