The answer is 20C6 which is 20!/[6!(20--6)!]
= 20*19*18*17*16*15/(6*5*4*3*2*1) = 38760
n! represents a*2*3*4*...*n
If their sequence matters . . . 120 If it doesn't . . . 20
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
120 5x4x3x2x1=120
If you can repeat the numbers, it will be 6*6*6=216. If you can't, it will be 6*5*4=120.
120 ABCDE , there are 120 ways these five letters can be rearranged.
There are: 10C3 = 120
There are 6!/3! = 120 different 3-digit numbers that can be made from these 6 digits.
720 permutations (the arrangement matters) 120 combinations (the arrangement doesn't matter)
120
120
The positive factors of 120 are 1, 2, 3, 4, 6, 10, 12, 20, 30, 40, 60, and 120. So that possible combinations of the length and width are, 1 and 120, 2 and 60, 3 and 40, 4 and 30, 6 and 20, and 10 and 12.
There are 120 permutations and 5 combinations.