There are 11C2 = 11*10/(2*1) = 55 combinations.
There are 11.There are 11.There are 11.There are 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?
1 and 11 and 21 and 31 and 42 and 12 and 22 and 32 and 43 and 13 and 23 and 33 and 44 and 14 and 24 and 34 and 416 combinations
11
There are 11C2 = 11*10/(2*1) = 55 combinations.
The number of combinations, denoted by 11C6 is 11!/[6!*(11-6)!] = 11*10*9*8*7/(5*4*3*2*1) = 462
There are several combinations of odd numbers that add up to 43. For example, 3 + 29 + 11 is equal to 43, so is 5 + 27 + 11, and many more. In fact, if you allow negative numbers, you have an infinite number of combinations.
There are 11.There are 11.There are 11.There are 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?
There are 24C12 = 24*23*...*13/(12*11*...*1) = 2,704,156 combinations.
1 and 11 and 21 and 31 and 42 and 12 and 22 and 32 and 43 and 13 and 23 and 33 and 44 and 14 and 24 and 34 and 416 combinations
You can get only four combinations: They are: 11, 118, 119 and 1189. In a combination, the order of the digits does not matter.
11
Assuming you meant how many combinations can be formed by picking 8 numbers from 56 numbers, we have:(56 * 55 * 54 * 53 * 52 * 51 * 50 * 49)/8! = (7 * 11 * 3 * 53 * 13 * 51 * 25 * 7) = 1420494075 combinations. (Also equal to 57274321104000/40320)
If you can use the same number twice, such as 11 or 22, then there are 100 possible numbers (00, and 01-99). If you can not use the same number twice, the answer is 90.
(13 x 12 x 11 x 10)/(4 x 3 x 2 x 1) = 715 of them