The challenge that Prof. Loh left us with at the end

Module 3 Week 1 Day 2 Your Turn Part 2
Prof. Loh left us with a little challenge at the end, but I can't seem to get what he meant. So far, I have worked out (10 x 30 squares x 30 squared)  30 squared is that what he meant?

@TheRogueBlade Wow, it's so awesome that you are trying this challenge out! I'm really excited that you asked about this. And it's so much fun to find a different way to solve a problem.
Prof. Loh says:
"If you want another challenge, to go even beyond here, see if you can find another way to get the same answer, as a \(\textcolor{red}{\text{product}},\) multiplying, where you first think about what happens with the tens, hundreds, and thousands, and multiply that by the number of ways to finish with the ones digit and the ten thousands digit."
Prof. Loh is telling us to
 Find the ways to choose the \( \textcolor{red}{\text{ middle three }} \) digits
 Find the ways to choose the \( \textcolor{blue}{\text{ first }} \) digit
 Find the ways to choose the \( \textcolor{purple}{\text{last}}\) digit
and multiply these three numbers together.
1.
For the middle three digits, they can take the form
$$ \textcolor{red}{B}\textcolor{blue}{CC}, $$
$$ \textcolor{red}{BB}\textcolor{blue}{C}, $$
$$ \text{ or } \textcolor{red}{BBB} $$
where \( \textcolor{red}{B} \neq \textcolor{blue}{C}\) and they are both single digits.
The important thing about the digit \( \textcolor{red}{B}\) is that it can be \(0,\) since we are considering the middle digits here.
So there are \(\textcolor{red}{10}\) choices for \(\textcolor{red}{B}.\) Since \(\textcolor{blue}{C}\) has to be different, there are \(\textcolor{blue}{9}\) choices for \(\textcolor{blue}{C}.\)
$$\begin{aligned} \text{ Ways to choose a value for } \textcolor{red}{B} \text{ and a value for } \textcolor{blue}{C} &= \textcolor{red}{10} \times \textcolor{blue}{9} = 90. \\ \end{aligned} $$Once we have chosen \(\textcolor{red}{B} \text{ and } C,\) we can get two numbers out of them: \(\textcolor{red}{B}CC \text{ and } \textcolor{red}{BB}C.\) So for each value of \( (\textcolor{red}{B}, \textcolor{blue}{C}) \) we get two numbers. Thus the \(90\) ways to choose \( ( \textcolor{red}{B}, \textcolor{blue}{C}) \) give \(180\) actual threedigit numbers.
(For example, maybe we choose \(\textcolor{red}{B} = 6 \text{ and } \textcolor{blue}{C} = 5. \) This gives us the two numbers of \(655\) and \(665.\) )
We mustn't forget about the ways that are like \( \textcolor{red}{BBB}.\) There are \(\textcolor{red}{10}\) values of \(\textcolor{red}{B},\) so \(10\) numbers of the form \( \textcolor{red}{BBB}.\)
$$\begin{aligned} \text{ Ways for } \textcolor{red}{B}\textcolor{blue}{CC}, \textcolor{red}{BB}\textcolor{blue}{C}, \text{ and } \textcolor{red}{BBB} &= 180 + 10 \\ &= \boxed{190} \\ \end{aligned} $$2.
There are \(9\) choices for the first digit, since the first digit can't be \(0.\)
3.
There are \(10\) choices for the last digit, which can be the same as \(\textcolor{red}{B}\) or \(\textcolor{blue}{C}.\)
Thus,
$$\begin{aligned} \text{ Ways to have hundreds = tens \textcolor{purple}{\text{ or } } hundreds = thousands } &= \left( \text{ ways for middle three } \right) \times \left( \text{ ways for first}\right) \times \left( \text{ ways for last } \right) \\ &= 190 \times 9 \times 10 \\ &= 1710 \times 10 \\ &= \boxed{17100}\\ \end{aligned} $$