How many $5$ digit natural numbers have at least one $5$ or one $7$? There are $9 \cdot 10^4=90,000$ possible $5$ digit numbers.
Numbers with no $5$ is $8\cdot 9\cdot 9\cdot 9\cdot 9 = 52,488$.
Numbers with no $7$ is $52,488$.
Numbers with at least one $5$ is $90,000 - 52,488$ 
Similarly numbers with at least one $7$ is $90,000-52,488$
So answer is $2(37,512)=75,024$
But this answer is wrong. Please provide correct solution..

Hint: Count the number of five digit-numbers that contain NO $5$'s and  no $7$'s; that gives us  $7\cdot 8^4$ such five-digit numbers.
The remainder ($90000 - 7\cdot 8^4$) will be the number of five digit numbers with at least one $5$ OR one $7$.
