A 333-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 333-digit numbers are there?


Answer:

285285285

Step by Step Explanation:
  1. A 333digit number is of the form ‘high-low-high’, so, the tens digit of the 333-digit number cannot be 9,9,9, as the units and the hundreds, digit needs to be larger than tens digit and 999 is the largest digit. Therefore, the smallest tens digit of the 333-digit number of the required form is 000 and the largest tens digit of the 333-digit number is 8.8.8.
  2. If the tens digit is 0,0,0, then the hundreds digit can be any digit from 111 to 9,9,9, and the units digit can also be any digit from 111 to 9.9.9.
    So, there are 9×99×99×9 possible numbers of the required form with 000 at tens place.
    If the tens digit is 1,1,1, then the hundreds digit can be any digit from 222 to 9,9,9, and the units digit can also be any digit from 222 to 9.9.9.
    So, there are 8×88×88×8 possible numbers of the required form with 111 at tens place.
  3. Similarly, possible numbers with 222 at tens place is 7×7,7×7,7×7, possible numbers with 333 at tens place is 6×6,,6×6,,6×6,, possible numbers with 888 at tens place is 1×1.1×1.1×1.
    Therefore, the total number of possible 333-digit numbers of the required form =(9×9)+(8×8)++(1×1)=285=(9×9)+(8×8)++(1×1)=285=(9×9)+(8×8)++(1×1)=285
  4. Hence, there are 285 3285 3285 3-digit numbers of the form ‘high-low-high’.

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