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Can some one please help me with this?

How many 10 digit phone numbers have at least one of each odd digit?

$1,3,5,7,9 = 5!*10^5$

I hope this is right. Or did I double count?

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    I don't think you've counted enough. I believe $5! 10^5$ would give the number of 10 digit phone numbers whose first 5 digits correspond to $1,3,5,7,9$ (in some order) and whose final 5 digits are anything. This would miss numbers like $1111113579$.2012-02-22
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    Your procedure does not give the right answer for much smaller cases (like $3$-digit numbers, from digits $1$, $2$, $3$, at least one of each odd digit).2012-02-22
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    But of course, ten digit phone numbers as they appear in the US have more constraints, such as not beginning with 0 or 911 and a whole slew of similar restrictions (which may be getting fewer, as new area codes come into existence). But then we are of course out of the realm of mathematics …2012-02-22

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