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.

Guest May 18, 2023

edited by
Guest
May 18, 2023

#1**0 **

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.**

Guest May 19, 2023