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 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.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
If the digits can repeat, then there are 256 possible combinations. If they can't repeat, then there are 24 possibilities.