Assuming no repeated digits, lowest first, 20; in any order 120;
Allowing repeated digits: 216
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.
The short answer is 1000. This is very easy to visualise: Simply consider each number in the combination to be a digit in a decimal number. We then end up with a three-digit number. Such a three-digit number ranges in value from 000 to 999, or 1000 unique combinations.
The order of the digits in a combination does not matter. So 123 is the same as 132 or 312 etc. There are 10 combinations using just one of the digits (3 times). There are 90 combinations using 2 digits (1 once and 1 twice). There are 120 combinations using three different digit. 220 in all.
There are 28706 such combinations. 5456 of these comprise three 2-digit numbers, 19008 comprise two 2-digit numbers and two 1-digit numbers, 4158 comprise one 2-digit number and four 1-digit numbers and 84 comprise six 1-digit numbers.
Using the word "combinations" in the English sense (as opposed to mathematical sense the expert has used) where it often used in the mathematical sense of the word "permutations": Assuming the hundreds digit must be at least 1 (eg 99 = 099 is not considered a three digit number), then: 9 x 10 x 10 = 900.
To find the number of three-digit combinations, we consider the digits from 000 to 999. Each digit can range from 0 to 9, giving us 10 options for each of the three digits. Therefore, the total number of three-digit combinations is (10 \times 10 \times 10 = 1,000).
1234, 1235 and 1236.
There are a total of 1,000 three-digit combinations from 000 to 999. This includes all combinations where the digits can range from 0 to 9, allowing for repetitions. Each of the three digit positions can have 10 possible values (0-9), leading to (10 \times 10 \times 10 = 1,000) combinations.
It depends on how many digit you are choosing from.
5
To form a three-digit number using the digits 1-7, we can choose any of the 7 digits for each of the three places (hundreds, tens, and units). Therefore, the total number of 3-digit combinations can be calculated as (7 \times 7 \times 7), which equals 343. Thus, there are 343 different three-digit numbers that can be formed using the digits 1-7.
To form three-digit even numbers from the set {2, 3, 5, 6, 7}, we can use the digits 2 or 6 as the last digit (to ensure the number is even). For each case, we can choose the first two digits from the remaining four digits. For three-digit numbers, there are 2 options for the last digit and (4 \times 3 = 12) combinations for the first two digits, resulting in (2 \times 12 = 24) even numbers. For four-digit even numbers, we again have 2 options for the last digit. The first three digits can be selected from the remaining four digits, giving us (4 \times 3 \times 2 = 24) combinations for each last digit. Thus, there are (2 \times 24 = 48) even four-digit numbers. In total, there are (24 + 48 = 72) three-digit and four-digit even numbers that can be formed from the set.
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.
One hundred is the least three-digit square number. It is formed by 102.
9
512
The first digit has 4 choices for its digit. The second digit has 6 choices and the third has 3. The solution would simply be 4*6*3=72 three digit numbers.