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27
3
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How many three-digit positive integers exist, all of whose digits are 2's and/or 5's and or 8's?

 Jan 3, 2021
 #1
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+1

Total Permutations =3^3 = 27

 

{2, 2, 2} | {2, 2, 5} | {2, 2, 8} | {2, 5, 2} | {2, 5, 5} | {2, 5, 8} | {2, 8, 2} | {2, 8, 5} | {2, 8, 8} | {5, 2, 2} | {5, 2, 5} | {5, 2, 8} | {5, 5, 2} | {5, 5, 5} | {5, 5, 8} | {5, 8, 2} | {5, 8, 5} | {5, 8, 8} | {8, 2, 2} | {8, 2, 5} | {8, 2, 8} | {8, 5, 2} | {8, 5, 5} | {8, 5, 8} | {8, 8, 2} | {8, 8, 5} | {8, 8, 8} (total: 27)

 Jan 3, 2021
 #2
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If you decide to list all of them,

You would get 27(As Guest said)

 

But there is an easier way, 

The first digit has three possibilities; it can either be a 2 a 5 or an 8. 

 

The second digit has three possibilities; it can either be a 2 a 5 or an 8.

 

The third digit has three possibilities; it can either be a 2 an 8 or a 5.

 

So \(3*3*3=27 \) possible integers.

 

cheekycheekycheeky

 Jan 3, 2021
edited by DewdropDancer  Jan 3, 2021
 #3
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0

I edited it, so it should be right now!

cheekycheekycheeky

DewdropDancer  Jan 3, 2021
edited by DewdropDancer  Jan 3, 2021

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