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
There are: 10C3 = 120
120 ABCDE , there are 120 ways these five letters can be rearranged.
If you can repeat the numbers, it will be 6*6*6=216. If you can't, it will be 6*5*4=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
There are 120 permutations and 5 combinations.
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.