$\begin{array}{1 1}(A)\;70\\(B)\;80\\(C)\;90\\(D)\;60\end{array} $

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

Let 2 be fixed at unit's place.Now we have 5 digits and 3 places are to be filled up.

This can be alone in $5P_2$ ways.

When unit's place is filled up both 4 or 6 again in each case we have $5P_3$ numbers

$\therefore$ 4-digit even numbers are

$3\times 5P_2=3\times \large\frac{5!}{(5-2)!}$

$\Rightarrow 3\times \large\frac{5!}{3!}$

$\Rightarrow 3\times \large\frac{5\times 4\times 3\times 2\times 1}{3\times 2\times 1}$

$\Rightarrow 3\times 5\times 4$

$\Rightarrow 60$

Hence (D) is the correct answer.

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