How many $5$ digit natural numbers have at least one $5$ or one $7$?

Question

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..

You've double-counted those numbers which have both a 7 and a 9. — Dustan Levenstein, Apr 01 '15 at 15:15
Count the "bad" numbers, neither $5$ nor $7$. Subtract from $9\times 10^4$. — André Nicolas, Apr 01 '15 at 15:18
You counted numbers that have both a $5$ and a $7$ twice. (For example, the number $10057$ is counted twice.) — Akiva Weinberger, Apr 01 '15 at 15:19

Jordan Glen · Answer 1 · 2015-04-01T15:21:23.593

4

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 $7$.

edited Apr 01 '15 at 15:21

answered Apr 01 '15 at 15:15

Jordan Glen

1 Answers1