Assuming no repetition, 12 x 11 x 10.... x 4 x 3 x 2.
If repetition is allowed, 1212
There are 24C12 = 24!/[12!*12!] = 2,704,156 combinations.
12
There are 24C12 = 24*23*...*13/(12*11*...*1) = 2,704,156 combinations.
12
30
Only three: 12, 13 and 23. Remember that the combinations 12 and 21 are the same.
26 = 64 combinations, including the null combination - which contains no numbers.
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.
The number of combinations of 12 numbers taken 12 at a time (i.e., choosing all 12 numbers from a set of 12) is calculated using the binomial coefficient formula, which is ( \binom{n}{k} = \frac{n!}{k!(n-k)!} ). For ( n = 12 ) and ( k = 12 ), this simplifies to ( \binom{12}{12} = 1 ). Therefore, there is only one combination of 12 numbers from 1 to 12, which includes all the numbers themselves.
To calculate the number of 12-number combinations using numbers 1-36, we can use the formula for combinations: nCr = n! / r!(n-r)!, where n is the total number of items to choose from (36) and r is the number of items to choose (12). Plugging in the values, we get 36C12 = 36! / 12!(36-12)! = 36! / 12!24! = (363534*...25) / (121110...*1). This simplifies to 125,736,770 unique combinations.
11
There are 12C4 4 NUMBER combinations. And that equals 12*11*10*9/(4/3/2/1) = 495 combinations. However, some of these, although 4 number combinations consist of 7 digits eg 1, 10, 11, and 12. Are you really sure you want 4-DIGIT combinations?