Assuming digits may not be repeated, there are 10C4 = 10*9*8*7/(4*3*2*1) = 210 of them and I regret that I am not minded to go through the list.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
the answer is = first 2-digit number by using 48= 28,82 and in 3 digit is=282,228,822,822
There are twelve possible solutions using the rule you stated.
10,000 combinations.
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.
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.
To calculate the number of different 4-digit combinations that can be made using numbers 0 through 9, we use the concept of permutations. Since repetition is allowed, we use the formula for permutations with repetition, which is n^r, where n is the number of options for each digit (10 in this case) and r is the number of digits (4 in this case). Therefore, the number of different 4-digit combinations that can be made using numbers 0 through 9 is 10^4, which equals 10,000 combinations.
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.
Using the eight digits, 1 - 8 ,-- There are 40,320 eight-digit permutations.-- There is 1 eight-digit combination.
There is only 1 combination.
There are 840 4-digit combinations without repeating any digit in the combinations.
To calculate the number of 4-digit combinations you can get from the numbers 1, 2, 2, and 6, we need to consider that the number 2 is repeated. Therefore, the total number of combinations is calculated using the formula for permutations of a multiset, which is 4! / (2!1!1!) = 12. So, there are 12 unique 4-digit combinations that can be formed from the numbers 1, 2, 2, and 6.