How many 3 digit numbers contain the digit 3 at least once?

Answer 1

172

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Another contributor was amazed:

Wow ! That's a neat little problem.

Let's see . . . . .

-- There are three ways to have a single 3 in the number: 3xx, x3x, and xx3.

For 3xx, there can be 81 different numbers. (2nd and 3rd digits can't be a 3. 9x9=81.)

For x3x, there can be 72 numbers. (1st can't be 0 or 3. 3rd can't be a 3. 8x9=72.)

For xx3, there are also 72. (1st can't be 0 or 3. 2nd can't be a 3. 8x9=72.)

So far, we have 81+72+72 = 225 possibilities, with a single 3 in each one.

-- There are three ways to have two 3s in the number: 33x, 3x3, and x33.

For 33x, there can be 9 different numbers. (3rd digit can't be a 3.)

For 3x3, there can be 9 numbers. (2nd digit can't be a 3.)

For x33, there are 8 numbers. (1st digit can't be a 0 or a 3.)

So we have 9+9+8 = 26 more possibilities.

-- There is only one way to have three 3s in the number: 333.

That's one more possibility.

So the total number of 3-digit numbers where at least one of the digits is a 3 is

225 + 26 + 1 = 252 all together.