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.
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2 Answers
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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 $2,5,8$.
$10^5 - 3\cdot 9^5 + 3\cdot 8^5 - 7^5=4350$
answered Feb 15, 2019 at 23:01
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$\begingroup$ Thank you so much, I tried a similar procedure, maybe I made a calculing mistake. $\endgroup$– rik99Feb 15, 2019 at 23:09
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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: $N=2940+1260+60+90=4350$.
