### A $3$-digit number is of the form ‘high-low-high’ $-$ that is the tens digit is smaller than both the hundreds digit and the units (or ‘ones’) digit. How many such $3$-digit numbers are there?

$285$

Step by Step Explanation:
1. A $3-$digit number is of the form ‘high-low-high’, so, the tens digit of the $3$-digit number cannot be $9,$ as the units and the hundreds, digit needs to be larger than tens digit and $9$ is the largest digit. Therefore, the smallest tens digit of the $3$-digit number of the required form is $0$ and the largest tens digit of the $3$-digit number is $8.$
2. If the tens digit is $0,$ then the hundreds digit can be any digit from $1$ to $9,$ and the units digit can also be any digit from $1$ to $9.$
So, there are $9 \times 9$ possible numbers of the required form with $0$ at tens place.
If the tens digit is $1,$ then the hundreds digit can be any digit from $2$ to $9,$ and the units digit can also be any digit from $2$ to $9.$
So, there are $8 \times 8$ possible numbers of the required form with $1$ at tens place.
3. Similarly, possible numbers with $2$ at tens place is $7 \times 7,$ possible numbers with $3$ at tens place is $6 \times 6, \ldots ,$ possible numbers with $8$ at tens place is $1 \times 1.$
Therefore, the total number of possible $3$-digit numbers of the required form $= (9 \times 9) + (8 \times 8) + \ldots + (1 \times 1) = 285$
4. Hence, there are $285 \space 3$-digit numbers of the form ‘high-low-high’.