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

Step by Step Explanation:
  1. A 3digit 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 9×9 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.
  3. 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
  4. Hence, there are 285 3-digit numbers of the form ‘high-low-high’.

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