There exist three types of number cases to be solved.
i) _ _ 00
ii) 11_ _
iii) _00_
In case i)
For _ _00, The last 2 digits can be from 00,11,22,...99 => 10 possibilities
The first digit can be filled in 9 ways out of 10 numbers (0 to 9).
The second digit can be filled in 8 ways out of 10 numbers, in which 0 and the number filled in first digit's place should be excluded. So total 9∗8=72 ways
Now, consider _ _ 11, _ _ 22,...etc
The first digit can be filled in 8 ways as 0 can't be filled in the first digit and 1 also cannot be included.
the second digit can be filled in 8 ways, as 0 can be included now but the digits in first and last position of the number should be excluded. So a total of 8∗8=64 ways.
These 64 ways are common for _ _11 to _ _ 99. So 64∗9=576 ways
Total possible numbers for case i) are \(576+72=648\) numbers
In case ii)
The first two digits can be from 11 to 99, 9 ways.
The third digit can be filled in 9 ways and the fourth digit can be filled in 8 ways. So total ways are 9∗8=72 ways.
Total possible numbers are \(72∗9=648\) numbers.
In case iii)
For _00_, the first digit can be filled in 9 ways and the last digit can be filled in 8 ways. Total ways are 9∗8=72 ways
For _11_, _22_, .... _99_ , The first digit can be filled in 8 ways (excluding 0 and digit in second place) and the last digit can be filled in 8 ways (including 0 and excluding 2 other digits already in the number). So total ways are 8∗8=64 ways for each possibility and there are 9 such possibilities, so total ways are 64∗9=576 ways.
Total possible numbers are \(576+72=648\) ways.
Finally, the total possible numbers are \(648+648+648=1944\) numbers.
OPTION :
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