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To do these problems think of four boxes, where each box represents a digit.

In the first box, you can only have 1-9, excluding 7 (1, 2, 3, 4, 5, 6, 8, 9), that's 8 numbers. Basically, it's the same as the next three boxes, except zero is also excluded.

In the second, third, and forth boxes, you can have 0-9, excluding 7, which is 9 numbers.

To find the answer, multiply the possibilities by each other.

"Box 1" x "Box 2" x "Box 3" x "Box 4"

8 x 9 x 9 x 9 = 81 x 72 = 5832 possibilities

The answer is 5832 possibilities.

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Q: How many integers between 1000 and 9999 inclusive contain no 7's?
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