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 9C3 = 84 combinations.
There are 9C3 = 10*9*8/(3*2*1) = 120 of them.
To calculate the number of 4-digit combinations using the digits 1, 3, 5, and 7 exactly once each, we can use the permutation formula. There are 4 choices for the first digit, 3 choices for the second digit, 2 choices for the third digit, and 1 choice for the fourth digit. Therefore, the total number of combinations is 4 x 3 x 2 x 1 = 24. So, there are 24 possible 4-digit combinations using the digits 1, 3, 5, and 7 exactly once each.
It depends. If you can only use each number once, there are 720 combinations. If you can use numbers multiple times, then there are 1000 combinations, by using all numbers from 000 to 999.
45*44*43*42*41/(5*4*3*2*1) = 1,221,759
There are 15180 combinations.
I am assuming you mean 3-number combinations rather than 3 digit combinations. Otherwise you have to treat 21 as a 2-digit number and equate it to 1-and-2. There are 21C3 combinations = 21*20*19/(3*2*1) = 7980 combinations.
There are 9C3 = 84 combinations.
6 of them.
There are 27 possible combinations.
A 3-digit safe code can have combinations ranging from 000 to 999. This gives a total of 1,000 possible combinations, as each digit can be any number from 0 to 9. Therefore, the total number of combinations is 10 (choices for the first digit) × 10 (choices for the second digit) × 10 (choices for the third digit), which equals 1,000.
the answer is = first 2-digit number by using 48= 28,82 and in 3 digit is=282,228,822,822
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
To find the number of combinations of the digits 1, 2, 3, 4, 5, and 6 that form numbers less than 500, we can consider the constraints based on the first digit. If the first digit is 1, 2, or 3, all combinations of the remaining digits can be used. If the first digit is 4, only combinations that result in a two-digit number can be formed. The total combinations can be calculated based on these conditions, but generally, you can form various 1-digit, 2-digit, and 3-digit numbers, totaling around 120 distinct combinations.
Allowing repetitions, there are 9 combinations. Without repeated digits, there is only one combination of 3 digits from 3.
The number of four-digit combinations is 10,000 .Stick a '3' before each of them, and you have all the possible 5-digit combinations that start with 3.There are 10,000 of them. They run from 30,000 to 39,999 .
To find the number of 3-digit combinations using the digits 0 to 9 with repetition allowed, we consider that each digit can be any of the 10 digits (0-9). Since there are 3 positions in the combination, the total number of combinations is calculated as (10 \times 10 \times 10), which equals 1,000. Therefore, there are 1,000 possible 3-digit combinations.