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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.

 May 18, 2023
edited by Guest  May 18, 2023
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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.

 May 19, 2023

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