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How many 5-digit numbers have at least one 2 or one 3 among their digits?

 Dec 18, 2019
 #1
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61328.

 Dec 18, 2019
 #2
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+1
 Dec 18, 2019
 #3
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What makes you think you are wrong? You were actually correct! The other answer was wrong. Thank you!

Guest Dec 18, 2019
 #4
avatar+108734 
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Guest 1:  How did you get this answer

 

Guest 2:   Why do you think the other answer was wrong.

 Dec 19, 2019
 #5
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Hi Melody: I don't know how Guest #1 got that answer which is CORRECT.

 

You arrive at the answer as follows:

 

1- Exclude all 2's and 3's from 2nd position to the left which leaves you with 8 numbers(0, 1, 4,5, 6, 7, 8, 9).

2 - This exclusion applies to the 3rd, 4th and 5th position.

3 - Therefore you have: 8 x 8  x 8 x 8 =4096

4 - Since there are only 7 numbers {1, 4, 5, 6, 7, 8, 9} to contend with, because we are excluding all 20,000 and 30,000 numbers, then we have: 7 x 4096 =28,672 numbers that must be excluded from the total of 90,000 5-digit integers.

5 - 90,000 - 28,672 = 61,328 numbers that must have either a 2 or a 3 or both in them.

6 - Just to confirm this, I wrote a short computer code to count them and got exactly this number of 61,328.

 Dec 19, 2019
 #6
avatar+108734 
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Thanks guest

 

I looked at my answer again and saw my error.

It is fixed now, with explanation, and the two answers do agree.

 

Here is the answer with the explanation.

https://web2.0calc.com/questions/help-please_56105

Melody  Dec 19, 2019
edited by Melody  Dec 19, 2019

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