In case anyone is wondering how this way compares with the method that Prof. Loh used in the Day 2 Your Turn lesson, the difference is this: @RZ923 is adding the ways in these three sections to get the final total number of ways:

M3W1D2-y-part-2-forum-adding-separate-sections.png

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$$ 8100 + 900 + 8100 = 17100 $$

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On the other hand, Prof. Loh calculates the ways using Inclusion-Exclusion. In fact, he can be quoted around 0:18 as saying, "Let me first not tell you how much is in this little piece (pointing to the left-side crescent moon shape), where the tens equals the hundreds and the hundreds does not equal the thousands... I will instead calculate the whole piece (the whole circle)."

That's just what he does! He adds the two circles together, and subtracts the overcounted area in the middle.

M3W1D2-y-part-2-forum-inclusion-exclusion.png

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$$ 9000 + 9000 - 900 = 17100 $$

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Yay, we get the same answer! Math is consistent! ðŸ™‚

]]>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 \( \textcolor{purple}{\text{last}}\) digitand 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 three-digit 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,

$$\begin{aligned} \text{ Ways to have hundreds = tens \textcolor{purple}{\text{ or } } hundreds = thousands } &= \left( \text{ ways for middle three } \right) \times \left( \text{ ways for first}\right) \times \left( \text{ ways for last } \right) \\ &= 190 \times 9 \times 10 \\ &= 1710 \times 10 \\ &= \boxed{17100}\\ \end{aligned} $$ðŸ™‚

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