Assuming you are treating each number as a number and not as an individual unit, the numbers you can make from these digits are 899, 989 and 998.
017, 071, 107, 170, 701, 710. 6 combinations
9*9*9 = 729 using the digits 1 to 9 and 2*9 using 10 and another digit. 749 in all.
there are 899 whole numbers that have three digits.
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Six combinations: 123, 132, 213, 231, 312, 321
Assuming you are treating each number as a number and not as an individual unit, the numbers you can make from these digits are 899, 989 and 998.
How many digits are in each combination ? ? -- If one, then there are three combinations: {5}, {6}, and {7) . -- If two, then there are three combinations: {5, 6}, (5, 7), and {6, 7}. -- If three, then there is only one combination: {5, 6, 7}
017, 071, 107, 170, 701, 710. 6 combinations
We notice that you said "combinations of three digits". You didn't say "3-digit numbers".There are (10 x 10 x 10) = 1000 combinations of three digits.The only combinations of 3 digits that form numbers less than 15 are 000, 001, 002, . . .012, 013, and 014. There are 15 of those.So the remaining (1000 - 15) = 985 are all larger than 15.
If the numbers contain zeros, the total number of combinations is 10,000. You can work this out easily logically: For ten single-digit numbers (0,1,2,3,4,5,6,7,8,9) then there are 10 possible 'combinations' For numbers with 2 digits then for each possible digit in the 10s column (e.g. in the 20s range) there are another 10 possible combinations (20,21,22,23,24,25,26,27, 28,29). As there are 10 possible ranges (single digits, teens, twenties, thirties etc) there will be 10 X 10 or 100 possible combinations. using the same logic, for three digits, there will be 10 X 10 X 10 or 1000 digits. And for 4 digits there will be 10 x 10 x 10 x 10 = 10,000 possible combinations. So for a number, say, with x digits, the total number of combinations of those digits will be 10 x 10 x 10..... etc with x numbers of 10s in the calculation. You can find out the number of combinations of any set of letters or numbers in the same way. as an example, to find out, say, the possible combinations of letters in the alphabet of 26 letters, then using the same method this can be given as 26 x 26 x 26 x 26............. with 26 '26's' in a row multiplied together. This gives the staggering amount of approximately 615612 followed by 31 zeros.
9*9*9 = 729 using the digits 1 to 9 and 2*9 using 10 and another digit. 749 in all.
576
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
there are 899 whole numbers that have three digits.
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