Statistics complement rule - help understand solution

[MaryStew]

What is the number of dec-strings d1,…,dn of length n such that d1d2≠00. In other words, what is the number of dec-strings of length n that don't begin with 00?

So the solution provided in the textbook states that its 99*(10^(n-2)) which I don't quite understand.

I know that the number of dec-strings of length n is 10^n. And I understand that to calculate the possibilities for the remaining digits past the first two, it'd be 10^(n-2). But I don't understand where the 99 came from.

Should it not be 9 * 9 * (10^(n-2)) since there's 9 choices for the first digit {1, 2,...9) and 9 choices for the second digit as well since it just can't start with 0? I'm confused as to where 99 came from and would appreciate some clarification.

 

[Harry_the_cat]

There are 100 ways of writing 2 digits: 00, 01, 02, ... , 99.
If you exclude 00, then there are 100-1=99 ways.
Initially you said it can't start with 00. But can it start with 01, 02 ... or 10, 20, ... ?
Your answer including 9x9 implies that neither of the first two digits can be 0.
Does d1d2 not equal 00 mean d1 x d2 not equal 0? Or are we talking about concatenation rather multiplication?

 

[Romsek]

there are 99 sequences of 2 decimal digits that aren't 00.

Pick one, that's 99 choices.

You now have [MATH]n-2[/MATH] bits left that can be anything. There are \(\displaystyle 10^{n-2}\) of those.

Hence a total of \(\displaystyle 99\cdot 10^{n-2}\)