How many ways can we select 14 numbers if we want to use each even digit at least once and the numbers can be repeated?
I figured that, we should take the maximum number of possibilites and extract when there is no even number.
The numbers are: 2,3,4,5,8,9
Maximum: $14^6$
There are 3 odd numbers so $14^3$
Final: $14^6 - 14^3 $
Is this correct? The selection order matters!
First of all the full number of ordered combinations is $6^{14}$, not $14^6$. From this number you should subtract all combinations which do not contain numbers $2,4,8$. Doing so we will however subtract twice every combination which does not contain two of these numbers and should correct for this adding the number of such combinations. Continuing further in this way (keyword - inclusion-exclusion principle) one obtains the answer: $$ 6^{14}-3\cdot5^{14}+3\cdot 4^{14}-3^{14}. $$