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?
Answer:
285
- 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.
- 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 9×9 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 8×8 possible numbers of the required form with 1 at tens place. - Similarly, possible numbers with 2 at tens place is 7×7, possible numbers with 3 at tens place is 6×6,…, possible numbers with 8 at tens place is 1×1.
Therefore, the total number of possible 3-digit numbers of the required form =(9×9)+(8×8)+…+(1×1)=285 - Hence, there are 285 3-digit numbers of the form ‘high-low-high’.