[Math] How many positive, three-digit integers contain at least one $3$

combinatorics

How many positive, three-digit integers contain at least one $3$ as a digit but do not contain $5$ as a digit?

I have an answer for that which is $215$ ,is that right ?

If its wrong then ,how to solve it?

Best Answer

I think your number is a little high. Take the number of 3-digit integers that contain no 5 and subtract the number of 3-digit numbers that contain no 5 and no 3.

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