Since all of the answers are at least 52, don't start counting! If that's the only method you can think of, take a guess and move on.

Since there are 9 groups of 100 among the three-digit integers from 100 to 999, it's best to group the possibilities into those sets of one hundred. For instance, consider the set 100 - 199. There are only two numbers that contain the digits 2 and 6. Since the hundreds digit is already occupied by 1, the tens and units digit must be 2 and 6 or 6 and 2. The possible numbers are 126 and 162.

The same rationale applies to the 300s, 400s, 500s, 700s, 800s, and 900s. That's two numbers each in seven 100-number groups, for a total of 14 possibilities.

That leaves us with the 200s and the 600s. Since those 100-number sets already contain a 2 or a 6 in the hundreds place, it stands to reason there will be more numbers with 2 and 6 as digits.

Consider the 200s. A number that contains both a 2 and a 6 must be of the form 2x6 or 26x. There are 10 numbers of the form 2x6: 206, 216, 226, and so on. There are also 10 numbers of the form 26x: 260, 261, 262, and so on. That might seem like there are 20 total possibilities in this set, but not so fast! The number 266 appears in both of those groups of 10, so it has been double-counted. There are really only 19 numbers in the 200s that also contain a 6.

The same reasoning applies to the 600s. Instead of 2x6 and 26x, we're looking at 6x2 and 62x. There are 19 possible numbers in this group as well.

The total, then, is 14 + 19 + 19 = 52, choice (A).