Using the digits 1 through 9, how many four-digit numbers can be formed in which 3 appears at least once?
The answer to this question is 2465. I would like to know the work involved in achieving this answer though.
You have 9 digits to start with. I assume you can repeat the digits. if that is the case, then:
You should have: 9 x 9 x 9 x 9 ==9^4 ==6,561 4-digit integers which include 3 in them.
If we remove the 3, then we have only 8 digits left, since we cannot start an integer with "0".
Therefore, we have: 8 x 8 x 8 x 8 ==8^4 ==4,096 - 4-digit integers which DO NOT include 3.
The difference: 6,561 - 4,096 ==2,465 4-digit integers that include at least one 3 in them.
