$\begin{array}{1 1}(A)\;191\\(B)\;192\\(C)\;193\\(D)\;194\end{array} $

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- $n!=n(n-1)(n-2)(n-3).....(3)(2)(1)$
- $nP_r=\large\frac{n!}{(n-r)!}$

Digits are 3,5,7,8,9

Besides 4 digits integers greater than 7000 five digits integers are always greater than 7000

The number of such integers is $5P_5=5!$

$5!=5\times 4\times 3\times 2\times 1$

$\Rightarrow 120$ ways

For a four digit integer to be greater than 7000 it must begin with 7,8 or 9.The number of such integer is $3\times 4P_3$

$4P_3=\large\frac{4!}{(4-3)!}=\frac{4!}{1!}$

$\Rightarrow 4\times 3\times 2\times 1$

$\Rightarrow 24$

$3\times 4P_3=3(24)$

$\Rightarrow 72$

Total no of ways =$120+72=192$

Hence (B) is the correct answer.

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