How many strings of three digits are there? 000 to 999, or a total of 1000.
How many strings of three digits contain the same three digits? That's 000, 111, 222 ... 999! ten in total.
The difference is your answer: 1000-10 = 990.
The number of strings of four decimal digits that have exactly three digits which are nine is the number of ways to choose three places to choose the nines multiplied by the number of possibilities for the other digits (because you can put all of the possibilities in an array so that each column has a different way of arranging the nines and that each row has a different set of other digits (because they do not affect each other)). The number of ways to choose the three places to choose the nines is the same as the number of ways of choosing 4-3=1 spot for there not to be a nine. The number of ways of choosing 1 spot where there are 4 spots is 4 (the first, the second, the third, or the fourth). Since there are 9 digits that are not nine, the number of choices for the other digits is 9 (since there is only one digit). Thus, there are 4*9=36 strings of four decimal digits that have exactly three digits which are nine.
Probably 990, depending on the rules. Fun question, but there is more than one answer depending on the rules. If the first digit can be zero, and if you can accept 001, 002, etcetera as answers, then you have 10 choices for the first digit, 10 choices for the second digit, and 10 choices for the third digit. But, you have to subtract the ten occurrences when all three digits match (000, 111, 222, 333, etc.) 10 x 10 x 10 - 10 = 990.
There are 625 of them - too many to list.
The answer depends on the degree of precision. If only integers are to be represented, then 6 digits would be enough because 165 = 1,048,576 is bigger than a million.
There are 256 bytes (8-bit strings), corresponding to the decimal numberszero through 255.Only the byte representing zero is composed of all zeros. All of the remaining255 bytes have at least one ' 1 '.
There are 5000 such strings.
There are 27 of them.
There can be only one.
9630 of them. 10 have all four digits the same and 360 have 3 of one and 1 of another.
There are 500000 such numbers.
One: 444!
There are 500 of 3-digit numbers begin with an odd one (100-199, 300-301, 500-599,700-799, and 900-999). In terms of "how many strings of", may filter or expand the above answer.
Only 1 exists, and it is "999"
If the lead digit cannot be zero then there are 9 x 9 x 8 x 7 = 4536 combinations. If the lead digit can be zero then there are, 10 x 9 x 8 x 7 = 5040 combination.
Billions are integers and so there will be no decimal points or digits after it.
The answer depends on how many decimal places are in the summands.
It will contain at least 12 digits