$\begin{array}{1 1}(A)\;360\\(B)\;125\\(C)\;150\\(D)\;175\end{array} $

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We have to look for no which ends with 2,4 or 6

The last digit has to end with 2,4,6

The other three digit no.can be formed =1,3,4,5,6,7 as the last digit is 2

Hence the no. of ways =$6P_3$

$\Rightarrow \large\frac{6!}{3!}$

$\Rightarrow \large\frac{6\times 5\times 4\times 3!}{3!}$

$\Rightarrow 120$

$\therefore$ Applying the same

We get the same for numbers ending with 4 or 6

Thus the total no of ways =$120\times 3$

$\Rightarrow 360$

Hence (A) is the correct answer.

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