How many numbers are there between $100$ and $1000$ such that every digit is either $2$ or $9$?

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How many numbers are there between $100$ and $1000$ such that every digit is either $2$ or $9$?
I couldn't understand what actually the question means. Does it talk about the numbers like $222$, $999$?
combinatorics permutations
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up vote
0
down vote
favorite
How many numbers are there between $100$ and $1000$ such that every digit is either $2$ or $9$?
I couldn't understand what actually the question means. Does it talk about the numbers like $222$, $999$?
combinatorics permutations
2
Also $229$, $929$, and so on.
â saulspatz
Aug 7 at 16:12
3
So, all the numbers between $100$ and $1000$ have 3 digits, right? So what can the first digit be? What can the second and the third?
â stuart stevenson
Aug 7 at 16:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How many numbers are there between $100$ and $1000$ such that every digit is either $2$ or $9$?
I couldn't understand what actually the question means. Does it talk about the numbers like $222$, $999$?
combinatorics permutations
How many numbers are there between $100$ and $1000$ such that every digit is either $2$ or $9$?
I couldn't understand what actually the question means. Does it talk about the numbers like $222$, $999$?
combinatorics permutations
asked Aug 7 at 16:10
blue_eyed_...
3,19321136
3,19321136
2
Also $229$, $929$, and so on.
â saulspatz
Aug 7 at 16:12
3
So, all the numbers between $100$ and $1000$ have 3 digits, right? So what can the first digit be? What can the second and the third?
â stuart stevenson
Aug 7 at 16:12
add a comment |Â
2
Also $229$, $929$, and so on.
â saulspatz
Aug 7 at 16:12
3
So, all the numbers between $100$ and $1000$ have 3 digits, right? So what can the first digit be? What can the second and the third?
â stuart stevenson
Aug 7 at 16:12
2
2
Also $229$, $929$, and so on.
â saulspatz
Aug 7 at 16:12
Also $229$, $929$, and so on.
â saulspatz
Aug 7 at 16:12
3
3
So, all the numbers between $100$ and $1000$ have 3 digits, right? So what can the first digit be? What can the second and the third?
â stuart stevenson
Aug 7 at 16:12
So, all the numbers between $100$ and $1000$ have 3 digits, right? So what can the first digit be? What can the second and the third?
â stuart stevenson
Aug 7 at 16:12
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
You have three digits. Each digit must be selected from the set $2,9$. You may select either of those two for each digit. So, the first digit has two choices. The second digit has two choices. The third digit has two choices. Each choice is independent of the previous choices. This allows you to apply the product principle.
There are $2cdot 2cdot 2=8$ such numbers:
222,
229,
292,
299,
922,
929,
992,
999
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
You have three digits. Each digit must be selected from the set $2,9$. You may select either of those two for each digit. So, the first digit has two choices. The second digit has two choices. The third digit has two choices. Each choice is independent of the previous choices. This allows you to apply the product principle.
There are $2cdot 2cdot 2=8$ such numbers:
222,
229,
292,
299,
922,
929,
992,
999
add a comment |Â
up vote
6
down vote
accepted
You have three digits. Each digit must be selected from the set $2,9$. You may select either of those two for each digit. So, the first digit has two choices. The second digit has two choices. The third digit has two choices. Each choice is independent of the previous choices. This allows you to apply the product principle.
There are $2cdot 2cdot 2=8$ such numbers:
222,
229,
292,
299,
922,
929,
992,
999
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
You have three digits. Each digit must be selected from the set $2,9$. You may select either of those two for each digit. So, the first digit has two choices. The second digit has two choices. The third digit has two choices. Each choice is independent of the previous choices. This allows you to apply the product principle.
There are $2cdot 2cdot 2=8$ such numbers:
222,
229,
292,
299,
922,
929,
992,
999
You have three digits. Each digit must be selected from the set $2,9$. You may select either of those two for each digit. So, the first digit has two choices. The second digit has two choices. The third digit has two choices. Each choice is independent of the previous choices. This allows you to apply the product principle.
There are $2cdot 2cdot 2=8$ such numbers:
222,
229,
292,
299,
922,
929,
992,
999
answered Aug 7 at 16:12
InterstellarProbe
2,262518
2,262518
add a comment |Â
add a comment |Â
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2
Also $229$, $929$, and so on.
â saulspatz
Aug 7 at 16:12
3
So, all the numbers between $100$ and $1000$ have 3 digits, right? So what can the first digit be? What can the second and the third?
â stuart stevenson
Aug 7 at 16:12