How many positive integers $< 1{,}000{,}000$ contain the digit $2$?

Though not always the smartest way, such questions can mechanically be answered as follows. (In this case the "smart" way to do it is Cameron's answer. It is instructive to see that this mechanical procedure basically recovers Cameron's method.) Let $a_n$ and $b_n$ be the amounts of $n$-digit numbers that do not and do have a $2$ in them. So $a_0=1$ and $b_0=0$. These number satisfy the recurrence $$ \begin{pmatrix}a_{n+1}\\b_{n+1}\end{pmatrix}= \begin{pmatrix}9&0\\1&10\end{pmatrix} \begin{pmatrix}a_n\\b_n\end{pmatrix} $$

(Take a moment to understand what this recurrence expresses.) Now $$ \begin{pmatrix}9&0\\1&10\end{pmatrix}^6 \begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}531441\\468559\end{pmatrix} $$

so the answer is $468559=10^6-9^6$.

The number of numbers from $1$ t0 $10^6$ that do not have the digit $2$ is clearly the same as the number of numbers that do not have the digit $9$. Now read each of these in base $9$ and you get all the numbers from 1 to $10^6$ (base 9) $=9^6$ (base 10). Therefore, there are $10^6-9^6$ numbers between $1$ and $10^6$ that use the digit 2.

I'm afraid you've miscounted. In this case, it would be better to count indirectly, by finding the numbers that don't have the digit $2$ in them, then subtracting these from the total.

First, let's count the number of $6$-digit numbers without a $2$ in them. There are $8$ choices for the leading digit of such a number, and for each of the other $5$ digits, there are $9$ choices. Thus, there are $8\cdot 9^5$ such numbers. Similarly, we can find that $8\cdot 9^4$ $5$-digit numbers without a $2$ in them, and so on, down to the $2$-digit numbers. Depending on whether $0$ is considered a $1$-digit number, there are either $8$ or $9$ numbers with one digit and no $2$'s. It turns out that the answer is not affected, either way, as I will discuss below.

Note Depending on whether you are taking $0$ to be a number, the number in the $1$-digit case will differ (though the answer, itself, will not). In fact, if you are taking $0$ to be a number, then the answer is greatly simplified, as you need only choose one of the $9$ available digits for each of the $6$ decimal places. This yields $9^6$ numbers less than $1000000$ without $2$ as a digit, out of a total of $1000000=10^6$ numbers less than $1000000.$ This also suggests an alternate approach in the case that $0$ is not a number being considered. Proceed as before, but discard zero as an option, so there are $9^6-1$ numbers less than $1000000$ without $2$ as a digit, out of a total of $999999=10^6-1$ numbers less than $1000000.$ In either case, there are $10^6-9^6$ numbers less than $1000000$ with $2$ as a digit.

This even agrees with the (more intuitive but less efficient) method outlined above. In general, we can find the sum using the formula for sums of geometric progressions. Alternately, here's a neat trick we can use.

Now, assume that $0$ is not among the numbers under consideration. (As we saw above, this won't make a difference.) In that case, there are $8=8\cdot 9^0$ single-digit numbers not equal to $2$. Hence, there are $$8\cdot9^5+8\cdot9^4+8\cdot9^3+8\cdot9^2+8\cdot9^1+8\cdot9^0$$ numbers less than $1000000$ that do not have $2$ as a digit. Let's call this sum $S$. Now, $$\begin{align}9S &= 9\left(8\cdot9^5+8\cdot9^4+ 8\cdot9^3+8\cdot9^2+8\cdot9^1+8\cdot9^0\right)\\ &= 8\cdot9^6+8\cdot9^5+8\cdot9^4+8\cdot9^3+8\cdot9^2+8\cdot9^1\\ &= 8\cdot9^6+S-8\cdot9^0\\ &= S+8\cdot\left(9^6-9^0\right)\end{align}$$ so $$8S=8\cdot\left(9^6-9^0\right),$$ and so $$S=9^6-9^0=9^6-1.$$ Since there are $10^6-1$ numbers less than $1000000,$ then as above, there are $$10^6-9^6=468559$$ numbers less than $1000000$ with $2$ as a digit.

You can get a generalised answer to this question (assuming that you are always asking how many integers with the digit 2 less than a particular power of 10).

For 10, there is 1
For 100, there is 1 + 1 + 10 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9x1 + 10 = 19
For 1000, there is 19 + 19 + 100 + 19 + 19 + 19 + 19 + 19 + 19 + 19 = 9x19 + 100 = 271

So, to generalise if n is the power of 10,

$$ A_1 = 1 $$

$$ A_n = 10^{n-1}+9A_{n-1} $$

So for 1,000,000,

$$ A_6=10^5+9A_5 $$ $$ A_6=10^5+9(10^4 + 9A_4) $$ $$ A_6=10^5+9*10^4+81*A_4 $$ $$ A_6=10^5+9*10^4+81(10^3+9A_3) $$ $$ A_6=10^5+9*10^4+81*10^3+729A_3 $$ $$ A_6=10^5+9*10^4+81*10^3+729(10^2+9A_2) $$ $$ A_6=10^5+9*10^4+81*10^3+729*10^2+6561A_2 $$ $$ A_6=10^5+9*10^4+81*10^3+729*10^2+6561(10+9A_1) $$ $$ A_6=100000+90000+81000+72900+65610+59049 $$ $$ A_6=468559 $$