The number of five digit numbers that contain 7 exactly once is

When 7 is right in the beginning, the number of numbers$= 9^4$ and when 7 is not in the beginning, the number of numbers is $\left({}^{4}C_1\right) \left(8\right) \left(9^3\right)$

$\therefore$ Required number of numbers

    $=9^4+\left({ }^4{C}_1\right)\left(8\right)9^3=\left(41\right)\left(9^3\right)$