using all digits from 0 to 9, allowing for repetition, how many 3 digit numbers are possible that contain at least one 5?

Let' first consider arrangements with just one 5.

As we require three digit number we consider three positions for the singe 5.

If we position it first, we can choose 9 digits for the second and 9 for the third

i.e. 1 x 9 x 9 => 81

If we position it second we have only 8 digits to choose for the first digit as we can't have zero or 5.

so, 8 x 1 x 9 => 72

Third we have, 8 x 9 x 1 => 72

With two 5's we have:

1 x 1 x 9

1 x 9 x 1

8 x 1 x 1

i.e. 26 in total

If all three digits are 5's we only have one way of doing this

Hence, in total 81 + 72 + 72 + 26 + 1 = 252 ways

Note: at least one 5 means all combinations - no 5's

No 5's => 8 x 9 x 9 = 648

Now, three digit numbers are from 100 to 999 inclusive

i.e. 900 numbers in total

And 900 - 648 = 252

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