If the 6 digits can be repeated, there are 1296 different combinations. If you cannot repeat digits in the combination there are 360 different combinations. * * * * * No. That is the number of PERMUTATIONS, not COMBINATIONS. If you have 6 different digits, you can make only 15 4-digit combinations from them.
about 1,0000000000000
10C6 = 10*9*8*7/(4*3*2*1) = 210 combinations.
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
If the 6 digits can be repeated, there are 1296 different combinations. If you cannot repeat digits in the combination there are 360 different combinations. * * * * * No. That is the number of PERMUTATIONS, not COMBINATIONS. If you have 6 different digits, you can make only 15 4-digit combinations from them.
Only one.
45 In combinations, the order of the digits does not matter so that 12 and 21 are considered the same.
about 1,0000000000000
9,000 - all the numbers between 1,000 and 9,999 inclusive. * * * * * NO. Those are PERMUTATIONS, not COMBINATIONS. Also, the question specified 4 digit combinations using 4 digits. The above answer uses 10 digits. If you start with 4 digits, you can make only 1 combination.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
10C6 = 10*9*8*7/(4*3*2*1) = 210 combinations.
Only one: 2468. The order of the digits in a combination does not make a difference.
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
There are 167960 9 digits combinations between numbers 1 and 20.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
There are infinite combinations that can make 3879