A $$3$$-digit integer is called a \textquotesingle V-number\textquotesingle{} if the digits go \textquotesingle high-low-high\textquotesingle{} $$-$$ that is, if the tens digit is smaller than both the hundreds digit and the units (or \textquotesingle ones\textquotesingle) digit.  How many $$3$$-digit \textquotesingle V-numbers\textquotesingle{} are there? 
Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Complex Forming Numbers
The smallest \textquotesingle V-number\textquotesingle{} is $$101$$ and the largest \textquotesingle V-mumber\textquotesingle{} is $$989$$.  Consider the tens digits. The smallest tens digit is $$0$$ and the largest tens digit is $$8$$.  If the tens digit is $$0$$, the hundreds digit can be $$1$$ to $$9$$, and the units digit can be $$1$$ to $$9$$, giving $$9 \times 9$$ possible \textquotesingle V-numbers\textquotesingle.  If the tens digit is $$1$$, then the hundreds digit can be $$2$$ to $$9$$ and the units digit can be $$2$$ to $$9$$, giving $$8 \times 8$$ possible \textquotesingle V-numbers\textquotesingle.  If the tens digit is $$d$$, where $$d$$ can be any digit from $$0$$ to $$8$$, the hundreds digit can be $$(d + 1)$$ to $$9$$ and the units digit can be $$(d + 1)$$ to $$9$$, giving $$(9-d)\times (9-d)$$ possible \textquotesingle V-numbers\textquotesingle.  The greatest value of $$d$$ is $$8$$. In this case, the hundreds digit can only be $$9$$ and the units digit can only be $$9$$, which gives just $$1 \times 1$$ possibilities.  This gives the total number of possible \textquotesingle V-numbers\textquotesingle{} to be $$9\times9+8\times8+\cdots +1\times1 = 285$$, which is the sum of the squares from $$1$$ to $$9$$ inclusive. 