Question 470481
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There are 5 even digits and 5 odd digits.


So if you allow repetition, there are 5 ways to select the first digit.  For each of those ways there are 5 ways to select the second digit.  For each of those 25 ways there are 5 ways to select the third digit.


So with repetition


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\ \times\ 5\ \times\ 5\ \times\ 5\ \times\ 5]


If repetion not allowed, there are 5 ways to select the first digit.  For each of those ways, since you don't allow repetition, there are only 4 ways to select the second digit, and then 3 ways to select the 3rd digit.  The 4th digit is now even, so we again have 5 ways, and the last is only 4 ways:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\ \times\ 4\ \times\ 3\ \times\ 5\ \times\ 4]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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