If no digit can be repeated then there are 5 combinations, abcd, abce, abde, acde and bcde. If you regard abdc as different from abcd then each of the 5 basic sets could be arranged 24 ways and the total would be 120 combinations.
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The question is incorrect since there are 210 - 1 possible combinations. The digit 0 can be in the combination or out. That gives 2 ways. With each, the digit 1 can be in the combination or out - 2*2 = 22 ways. With each, the digit 2 can be in the combination or out = 23 ways. With each, the digit 3 can be in the combination or out = 24 ways. etc. So 210 ways in all except that one of them is the null combination. Now 210 = 1024 so there are only 1023 combinations. If, instead, you allow the digits to be used many times, there is no limit to the number of combinations.
The only combination that we can be absolutely certain makes 24 is '2' and '4'. Any others depend on exactly how you plan to combine them arithmetically.
As no digit is used twice then you have 4 choices for the first digit, but only 3 choices for the 2nd digit, 2 for the 3rd and just one for the 4th. 4 x 3 x 2 x 1 = 24 combinations.
If no digit can be repeated then there are 5 combinations, abcd, abce, abde, acde and bcde. If you regard abdc as different from abcd then each of the 5 basic sets could be arranged 24 ways and the total would be 120 combinations.
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There are 9 1-digit numbers and 16-2 digit numbers. So a 5 digit combination is obtained as:Five 1-digit numbers and no 2-digit numbers: 126 combinationsThree 1-digit numbers and one 2-digit number: 1344 combinationsOne 1-digit numbers and two 2-digit numbers: 1080 combinationsThat makes a total of 2550 combinations. This scheme does not differentiate between {13, 24, 5} and {1, 2, 3, 4, 5}. Adjusting for that would complicate the calculation considerably and reduce the number of combinations.
If the digits can repeat, then there are 256 possible combinations. If they can't repeat, then there are 24 possibilities.
There are 24C12 = 24!/[12!*12!] = 2,704,156 combinations.
4!4!4!4!= 331776 with repetition Or 4!=24 without repetition.
There are 24C12 = 24*23*...*13/(12*11*...*1) = 2,704,156 combinations.
The question is incorrect since there are 210 - 1 possible combinations. The digit 0 can be in the combination or out. That gives 2 ways. With each, the digit 1 can be in the combination or out - 2*2 = 22 ways. With each, the digit 2 can be in the combination or out = 23 ways. With each, the digit 3 can be in the combination or out = 24 ways. etc. So 210 ways in all except that one of them is the null combination. Now 210 = 1024 so there are only 1023 combinations. If, instead, you allow the digits to be used many times, there is no limit to the number of combinations.
As no digit is used twice then you have 4 choices for the first digit, but only 3 choices for the 2nd digit, 2 for the 3rd and just one for the 4th. 4 x 3 x 2 x 1 = 24 combinations.
The only combination that we can be absolutely certain makes 24 is '2' and '4'. Any others depend on exactly how you plan to combine them arithmetically.
There are 24 combinations using each digit once per combination: 1234 1243 1324 1342 1423 1432 2134 2143 2314 2341 2413 2431 3124 3142 3214 3241 3412 3421 4123 4132 4213 4231 4312 and 4321.
With 4 digits you can have 24 = 16 combinations. 1 combination with 0 digits; 4 combinations with 1 digit: 1, 2, 4 and 8 6 combinations with 2 digits: 12, 14, 18, 24, 28 and 48 4 combinations with 3 digits: 124, 128, 148 and 248 1 combination with all 4 digits. In a combination the order of digits is not relevant so that 124 is the same as 142.