How many 6-digit numbers are there whose digits alternate between even and odd numbers? Just hoping to solve this question that asks how many 6-digit numbers there are, where each digit alternates between even and odd numbers? I know that you simply need to find the number of options each digit can hold but was unsure if I am supposed to add them up or multiply - given this is a mutually exclusive question. Thank you in advance!
 A: There are nine choices for the first digit, and five choices for each of the other five digits. The answer is therefore $9 \cdot 5^5=28125$.
A: There are $3$ even digits and $5$ choices they each can be.  There are $3$ odd digits and $5$ choices they each can be.
You can start with either an even digit or an odd digit.
If you start with an odd digit it can not be $0$ so there are $4$ choices.
That's enough.
Do not read the next paragraph:

  I told you not to read this!  Don't read the next one either.


 $5^3\cdot 5^3 + 4\cdot 5^2\cdot 5^3$

A: Imagine that you picked an even number for your first digit. In order to pick from even, you had 4 choices available (2, 4, 6, 8), since 0 is not a valid way to start a number. So the amount of 6 digit numbers whos digits alternate between even and odd and who begin with an even number are:
$4\cdot$ (the rest of the of the choices)
Why do we multiply here? Because for any 5 digit combination xxxxx that you pick, you get 4 more possibilities: 2xxxxx, 4xxxxx, 6xxxxx and 8xxxxx. This is why we multiply when there is an "interdependency" or the order matters.
Anyways, since your first digit was even, the next digit must be odd. In that case you have only 5 choices (1, 3, 5, 7, or 9).
So, so far you have:
$4\cdot5\cdot$(the rest of the choices)
Why do we multiply by 5 here? Again, given any 4 digit choice xxxx, you have 5 new possibilities by letting the previous digit vary (1xxxx, 3xxxx, 5xxxx, 7xxxx, 9xxxx), and for each of those possibilities you have another 4 possibilities by letting the first digit vary (21xxxx, 41xxxx, 61xxxx, 81xxxx, etc.). I hope this helps make clear that we will be multiplying these all together.
Now, since you picked an odd number for your first digit, you must pick an even number for the second one. This gives you another 5 choices. Likewise, your next number must be odd again - another five choices. We continue in this manner until we end up with:
$4\cdot5\cdot5\cdot5\cdot5\cdot5$
But remember! This is just one possible case. You assumed you started with an even digit, which is how you started off with 4. But you could also have started with an odd digit. In that case you would have 5, not 4 choices, since all 5 odd digits are acceptable beginnings to a number.
By the same reasoning as before, you have 5 choices for the next even number, 5 choices for the next odd number, etc. So in the case where you start with an odd number, there are:
$5\cdot5\cdot5\cdot5\cdot5\cdot5$
possibilities.
So now we have the number of possibilities for these two cases. Here's the crux: Do we add them or multiply them together to get the final answer?
Earlier, we multiplied numbers together, because for each choice, we were able to generate more possibilities. But in this case, the two paths are independent of each other. I had to choose at the start to do one or the other - I don't get to somehow go down both paths. So, since these two values are not dependent on each other, we add them together.
$4\cdot5\cdot5\cdot5\cdot5\cdot5 + 5\cdot5\cdot5\cdot5\cdot5\cdot5$
$ = (4 + 5)\cdot(5\cdot5\cdot5\cdot5\cdot5)$
$ = 9\cdot5^5 = 28125$
