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I made three case that $2$ is at extreme right then we have $9\times 10 = 90$ numbers, $2$ in middle has $81$ numbers,and with $2$ in beginning has $90$ numbers. This makes $90+90+81=261$ numbers and we add the numbers with no $2$ in them, that is $8\times 9 \times 9= 648$. Thus total will be $909$. What will be the common numbers counted multiple times and how to eliminate them?

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    Sometimes, when faced with problems of the type "how many ... such that some property does **not** hold?", I find it useful to try counting the number of possibilities which DO satisfy the property. I won't work it out here, but try thinking about how many 3-digit numbers are there which have a 1 adjacent to a 2 on the right side. Then you can subtract that from the total number of 3-digit numbers. – 2017-02-06

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Similar question is already asked here and also have answers.

Your method is also good but little lengthy.

So I tried your method to solve it also.

Your first mistake is when right most digit is 2 so it can choose in 1 way.

For first two digits you are doing 9 × 10.

But you forget in these you are still including numbers those have 2 on first place and 1 on second.

I want you to solve this type of mistake in each parts.

Still any doubt after solving please ask.

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    You answered the exact same question (with [the exact same answer](http://math.stackexchange.com/a/2130328/131263)) yesterday. Why not just point this out in a comment to the question??? – 2017-02-06
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    @barak manos I want to answer the users question according to his method also that's why. You want me to provide link to that answer and only answer according to user's method? Is that much better? – 2017-02-06
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    Dunno, but the question itself is most certainly a duplicate, and your answer looks rather similar to your answer on the previous question... But this is just a suggestion, so feel free to do whatever you think is more appropriate... – 2017-02-06
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    @Departed you solved it? Or having any doubt? – 2017-02-09
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The total number of 3 digit numbers are $9×10×10$

Now fix a number that starts with $21$ now there are $10$ possible choices for the unit's digit.So these numbers are 10 in number.

Now fix $21$ in the hundreds and units place respectively.Now you have only $9$ choices for the uits place.So these numbers are $9$ in numbers.

Hence you get total of $19$ numbers out of the total $900$ which violate the given condition.So just substractt 19 numbers.That gives you $881$