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So, I have this question i got from a math competition type of book, but it only has the numeric answer only. The question is:

What fraction of all four-digit natural numbers have a product of their digits that is even

The answer supposedly is : 8375/9000 == 67/72

I can understand how to get the denominator of this for the range of all possible 4 digit number from : 1000 -> 9999, the count is 9000.

But in terms of the numerator, I think one needs to go thru cases. Using the property that eveneven = even and oddeven = even. If just one digit is even of the 4 digits then the product of the digits will be even. SO we go thru the cases of one, then 2 then 3 then all 4 even numbers.

Just not sure how to get all the permutations/combinations here. SO need help here.

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    $\begingroup$ "SO we go thru the cases of one, then 2 then 3 then all 4 even numbers.".... Or we could subtract the cases where they are all odd. P(prod even) = P(at least 1 even) = 1 - P(all of them odd) = 1- (# ways all odd)/(# ways total) = 1 - (5^4)/9000. $\endgroup$ – fleablood Sep 3 '18 at 0:03
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As you say, if there is at least one even digit the product will be even, so count how many four digit numbers have all odd digits. There are five odd digits, so there are $5^4=625$ of them. The fraction of even products is then $\frac {9000-625}{9000}$

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  • $\begingroup$ Hi Ross, thanks for this. I totally forgot about think about Exclusion like principle approach. I will award this question to you. $\endgroup$ – Palu Sep 3 '18 at 0:08

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