5*4*3*2 = 120
The plural form is digits; the singular form is digit.
6 possible 3 digit combonations
Since there are no numbers and no underlined digits, there cannot be any answer!
To form a 5-digit number using the digits 1, 2, 6, 7, and 9 without repetition, we can use all five digits. The number of different arrangements of these 5 digits is calculated by finding the factorial of the number of digits, which is 5!. Therefore, the total number of 5-digit numbers that can be formed is 5! = 120.
To form a three-digit number using the digits 0-9, the first digit cannot be 0 (as it would not be a three-digit number). Thus, the first digit can be any of the digits from 1 to 9 (9 options). The second and third digits can each be any digit from 0 to 9 (10 options each). Therefore, the total number of three-digit numbers is (9 \times 10 \times 10 = 900).
You have seven different digits (symbols) to choose from, so you can form seven different one digit numbers and 7×7=72=49 different two digit numbers.
90 of them.
The answer will depend on how many digits you have at your disposal. Working with the 10 digits, and barring leading zeros, there are 8,877,690 numbers.
It depends upon whether the numbers can be used more than once. If the numbers can be used more than once, then there are 1,000 possible combinations; if not, then there are 720 possible combinations. ========== Assuming you are talking about integers you can calculate it this way: you can have any one of 9 digits for the first digit (zero is excluded because that would make it only a 2 digit number) You can have any one of 10 digits for the second and any one of 10 digits for the third digit. That gives you 9x10x10 = 900 different combinations for 3 digit numbers (not 1000). If you can include both negative and positive numbers as different numbers you get twice as many (2x900=18000). If you can count decimal numbers as 3 digit numbers (i.e. not restricted to integers) you can have 900 integers, 900 with the form xx.x, 1000 with the form x.xx (if zero is permitted to be the first digit and count as one of the 3 digits) or 900 (if a leading zero is NOT counted as one of the 3 digits). If a leading zero is NOT counted as one of the 3 digits, you could also have 1000 numbers of the form 0.xxx or just .xxx
6*5*4*3*2*1=720 possible numbers
The plural form is digits; the singular form is digit.
6 possible 3 digit combonations
The number of six digit numbers that you can make from ten different digits ifrepetitions of same digit on the six digit number is allowed is 1 000 000 numbers(including number 000 000).If no repetitions of the the same digit are allowed then you have:10P6 = 10!/(10-6)! = 151 200 different six digit numbers(six digit permutations form 10 different digits).
Since there are no numbers and no underlined digits, there cannot be any answer!
There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.
To form a three-digit number using the digits 0-9, the first digit cannot be 0 (as it would not be a three-digit number). Thus, the first digit can be any of the digits from 1 to 9 (9 options). The second and third digits can each be any digit from 0 to 9 (10 options each). Therefore, the total number of three-digit numbers is (9 \times 10 \times 10 = 900).
By using the 3 digits of a number we can form 3 different two digit numbers. 3C2 = 3!/[(3 - 2)!2!] = 3!/(1!2!) = (3 x 2!)/2! = 3