<PRE><B>Question 529127</B>
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6 ways to choose the first digit, then for each of those 6 ways there are 6 ways to choose the second digit, 6 times 6 = 36.  Then for each of those 36 ways to choose the second digit, there are 6 ways to choose the third digit.  36 times 6.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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