A 33-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 33-digit numbers are there?
Answer:
285285
- A 3−3−digit number is of the form ‘high-low-high’, so, the tens digit of the 33-digit number cannot be 9,9, as the units and the hundreds, digit needs to be larger than tens digit and 99 is the largest digit. Therefore, the smallest tens digit of the 33-digit number of the required form is 00 and the largest tens digit of the 33-digit number is 8.8.
- If the tens digit is 0,0, then the hundreds digit can be any digit from 11 to 9,9, and the units digit can also be any digit from 11 to 9.9.
So, there are 9×99×9 possible numbers of the required form with 00 at tens place.
If the tens digit is 1,1, then the hundreds digit can be any digit from 22 to 9,9, and the units digit can also be any digit from 22 to 9.9.
So, there are 8×88×8 possible numbers of the required form with 11 at tens place. - Similarly, possible numbers with 22 at tens place is 7×7,7×7, possible numbers with 33 at tens place is 6×6,…,6×6,…, possible numbers with 88 at tens place is 1×1.1×1.
Therefore, the total number of possible 33-digit numbers of the required form =(9×9)+(8×8)+…+(1×1)=285=(9×9)+(8×8)+…+(1×1)=285 - Hence, there are 285 3285 3-digit numbers of the form ‘high-low-high’.