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Answer C

The total number of three digit numbers is $9\times 10\times 10$ as the first place can be filled with $9$ ways $(1-9)$ and next two places can be filled with $10$ digits $(0-9)$ in $10 \times 10$ ways.

Now the total number of three digit number with $1$ immediate right of $2$ are

  • $(210-219) \implies 10$
  • $ \_$   $2$ $1 \implies$ this blank can be filled from $1-9 $ so total $ 9$ numbers here
  • Thus $10 +9 = 19$ numbers are having $1 $ to the immediate right of $2.$

So, total number of three digit numbers not having $1$ to the immediate right of $2 = 900 - 19 = 881$.

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Because 0 cannot come in the first place then 9 ways to fill first position, 10 ways for the next position and same way 10 ways for next position.

9*10*10=900 ways to have 3 digit numbers which dont start with 0

next there are 10 ways where first position is 1st  and 2nd position numbers are 1 and 2(121,122,123,124,125,126,127,128,129) since this is the only way to get position 1 and position 2 1 and 2 respectively then position 2 and 3 we can get in 9 ways 112,212,312,412,512,612,712,812,912 so sum of that is 19 so 900-19=881 ways.

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