How many numbers from 0 to 99999 contain the digits 2, 5 and 8?

Question

I have this problem:

How many integers between 0 and 99999 contain the digits 2, 5 and 8?

I've tried a lot, but I don't know how resolve it.

P.S. The solution should be 4350.

Answer 1

Consider the alternative question: How many numbers in that range Do Not contain at least one of $2,5,8$?

The number where we don't care is $10^5$

The number which do not contain $2$ is $9^5$. Similarly so for the number which do not contain $5$ as well as the number which do not contain $8$.

The number which do not contain $2$ as well as simultaneously not containing $5$ is $8^5$, etc...

Apply inclusion-exclusion and get a total number of numbers which do not contain at least one of $2,5,8$.

$10^5 - 3\cdot 9^5 + 3\cdot 8^5 - 7^5=4350$

Answer 2

Consider the cases, where $*$ is a digit different from $2,5,8$:

1) $258** \Rightarrow P(5,3)\cdot 7^2=2940$.

2) $2258*,2558*,2588* \Rightarrow 3\cdot \frac{P(5,4)}{2!}\cdot 7=1260.$

3) $22258,25558,25888 \Rightarrow 3\cdot \frac{P(5,5)}{3!}=60.$

4) $22558,22588,25588 \Rightarrow 3\cdot \frac{P(5,5)}{2!\cdot 2!}=90.$

Hence: $N=2940+1260+60+90=4350$.