Well in general, the pattern for all combinations of three digits A, B, C will be:
AAA, AAB, AAC, ABA, ABB, ABC, ACA, ACB, ACC, BAA, BAB, BAC, BBA, BBB, BBC, BCA, BCB, BCC, CAA, CAB, CAC, CBA, CBB, CBC, CCA, CCB, CCC
Yuo can make only one combination of 30 digits using 30 digits.
To calculate the number of 4-digit combinations that can be made with 4 digits, we can use the formula for permutations. Since there are 10 possible digits (0-9) for each of the 4 positions, the total number of combinations is 10^4, which equals 10,000. This is because each digit can be selected independently for each position, resulting in a total of 10 choices for each of the 4 positions.
If you have 24 distinct characters, then there are 2,704,156 combinations of 12 characters. Normally there are only ten digits and so 24 of them would contain duplicates. In that case the answer will depend on the duplication.
Only one: 2468. The order of the digits in a combination does not make a difference.
That depends on . . . -- how many digits you want in each combination -- whether it's allowed to repeat the same digit -- whether the order of the digits in the combination matters, i.e. whether you actually want combinations or permutations. Each choice has a different answer.
You can make: 1 combination containing 0 digits, 7 combinations containing 1 digits, 21 combinations containing 2 digits, 35 combinations containing 3 digits, 35 combinations containing 4 digits, 21 combinations containing 5 digits, 7 combinations containing 6 digits, and 1 combinations containing 7 digits. That makes 2^7 = 128 in all.
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
6 for 3-digits, 6 for 2-digits, 3 for 1-digits, and 15 for all of the combinations
Yuo can make only one combination of 30 digits using 30 digits.
If you use them only once each, you can make 15 combinations. 1 with all four digits, 4 with 3 digits, 6 with 2 digits, and 4 with 1 digit. There is also a combination containing no digits making 16 = 24 combinations from 4 elements.
9.
456, 546 and 564 are three of the four possible even numbers.
Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.
If you have 24 distinct characters, then there are 2,704,156 combinations of 12 characters. Normally there are only ten digits and so 24 of them would contain duplicates. In that case the answer will depend on the duplication.
You can make 27 unique three-digit numbers using the three digits 0, 1, and 2.i.e. for each of the three digits abc, there are 9 possible orders of those, from aaa to ccc.The 9 starting with 0:000001002010011012020021022* the combinations are the same as the numbers 0 to 26 expressed in base 3 (000 to 222)26 base 3 = 222 = 18 + 6 + 2 = 26
This question could would be best answered if you have a basic understanding of combinatorics (the study of combinations and permutations and probability) however, for something with as few possibilities as this, it is often easier to just make a list. Start with three digit combinations, then two digit combos, then one digit. Remember also that combinations are order independent ( ABC is the same thing as CBA) 3 digits 3 2 1 2 digits 3 2 3 1 2 1 1 digits 1 2 3 count those up and voila.
If digits can be reused, then 5x5x5x5x5 = 3125. If digits cannot be reused, then 5x4x3x2x1 = 120.