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How many 5-digit numbers have at least one 2 or one 3 or one 4 among their digits?

 Jan 7, 2021
 #1
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There are 10 different number 0-9. There are three digits and you want at least one to appear. That means the odds are 3/10.

 Jan 7, 2021
 #3
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the question didn't ask for the odds, it asked for the number of 5 digit numbers satisfying the statement.

 

Instead of counting the number of 5 digit numbers satisfying the statement, count the ones that don't satisfy the statement and subtract from the total number of possibilities.

 

For the first digit, there are 9-3=6 possibilities, and for the rest of the digits, there are 10-3=7 possibilities. Therefore the total of numbers that don't satisfy the statement is equal to 6*7*7*7*7 = 14406.

 

The total number of possibilities is just 9*10*10*10*10=90000.

 

Subtracting, we get the answer:

 

75594 5-digit numbers.

Guest Jan 7, 2021
 #2
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How many 5 digit numbers are there

99999-10000+1= 90000     or

9*10*10*10*10=90000

 

How many are there that do not have a 2 or 3 or 4 as any digit.

6*8^4 = 24576

 

So how many of them DO have at least one 2 or one 3 or one 4 

90000-24576 = 65424

 Jan 7, 2021
 #4
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I don't think you quite got it. There are 9*10^4 5 digit numbers and 6*7^4 that have no 2's, 3's, or 4's. Using complementary counting, the answer is 75594.

:D

 Jan 7, 2021

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