There are 8 possible numbers.
The last digit can be 1 or 7 since the number is odd.
The first digit can be any of the three except 0 since then it would not
be a 4 digit number, but once you pick the last digit, there are really only 2 numbers left for the first digit.
So let's use 1 as the last digit
the numbers are
7401, 7041, 4701, 4071.
Now use 7 as the last digit.
We have 1407, 1047, 4107, 4017.
So the 8 numbers are:
7401, 7041, 4701, 4071,1407, 1047, 4107, 4017
( the number of possible numbers is 8 since that is 23 . There are 2 choices for
the last digit, the once we pick that 2 choices for the first digit. For either of the remaining two digits there are 2 choices and then the last digit is fixed.)
1111, unless you allow negative number. In that case, the answer is -9999.
Total number of possible 3-digit numbers = 9!x10!10!
To form all possible three-digit numbers using the digits 0-9 without repetition, you can select any digit from 1-9 for the hundreds place (9 options), any digit from 0-9 for the tens place (10 options), and any remaining digit for the units place (8 options). However, since the first digit cannot be 0, the total number of valid combinations is calculated as 9 (choices for the first digit) × 9 (choices for the second digit, including 0) × 8 (choices for the third digit). This results in 9 × 9 × 8 = 648 unique three-digit numbers.
The total number of possible lock combinations depends on the number of digits and the range of each digit. For a standard 3-digit lock using numbers 0-9, there are 10 options for each digit, resulting in 1,000 possible combinations (000 to 999). If the lock has a different number of digits or uses letters, the calculation would vary accordingly. For example, a 4-digit lock would have 10,000 combinations.
10 because 10 is the first 2 digit number out of all numbers.
The sum is 154.
1111, unless you allow negative number. In that case, the answer is -9999.
1,000,000 or, if they need to be different, then 1,023,456.
Total number of possible 3-digit numbers = 9!x10!10!
Oh, what a happy little question! To find the smallest 6-digit number with no repeated digits, we start by using the digits 1-9 in ascending order. So, the smallest 6-digit number with no repeated digits is 123,456. Just imagine all the beautiful possibilities that number holds!
All nines 9,999,999,999
To solve this problem, we need to consider the possible combinations of a 4-digit number and a single-digit number using all the numbers from 1 to 9 exactly once. The only way to achieve this is by multiplying a 4-digit number by a single-digit number. One possible solution is 219 x 78 = 17082, where each digit from 1 to 9 is used exactly once.
To form all possible three-digit numbers using the digits 0-9 without repetition, you can select any digit from 1-9 for the hundreds place (9 options), any digit from 0-9 for the tens place (10 options), and any remaining digit for the units place (8 options). However, since the first digit cannot be 0, the total number of valid combinations is calculated as 9 (choices for the first digit) × 9 (choices for the second digit, including 0) × 8 (choices for the third digit). This results in 9 × 9 × 8 = 648 unique three-digit numbers.
The total number of possible lock combinations depends on the number of digits and the range of each digit. For a standard 3-digit lock using numbers 0-9, there are 10 options for each digit, resulting in 1,000 possible combinations (000 to 999). If the lock has a different number of digits or uses letters, the calculation would vary accordingly. For example, a 4-digit lock would have 10,000 combinations.
10 because 10 is the first 2 digit number out of all numbers.
All 9's.
5,040Assuming that the combination uses ALL single digits from 0 to 9, then for the first digit you will be able to use all 10 numbers, for the second digit you will be able to use 9, for the third digit 8 and for the last digit 7, giving a total number of combinations of 10 x 9 x 8 x 7 = 5,040 without the same number being used more than once in each combination.