Question 1169829
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Assuming that zero is disallowed for the high order digit, there are 5 ways to choose that digit, then 6 ways to choose the second digit, 6 ways to choose the third digit, and since there are 3 odd digits, 3 ways to choose the fourth digit.  In sum:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  5\,\times\,6\,\times\,6\,\times\,3]

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
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