A 5-digit combination can be obtained as:
There are 9C5 = 126 of the first kind,
9C3*30C1 = 84*30 = 2520 of the second, and 9C1*30C2 = 9*435 = 3915 of the third.
All together, there are 126 + 2520 + 3915 = 6561 5-digit combinations.
about 1,0000000000000
Oh, dude, you're hitting me with some math vibes here. So, if you have 6 digits to choose from to make a 4-digit combination, you can calculate that by using the formula for permutations: 6P4, which equals 360. So, like, you can make 360 different 4-digit combinations from those 6 digits. Math is wild, man.
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.
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.
Oh, dude, you're hitting me with some math vibes here. So, if you have 6 digits to choose from to make a 4-digit combination, you can calculate that by using the formula for permutations: 6P4, which equals 360. So, like, you can make 360 different 4-digit combinations from those 6 digits. Math is wild, man.
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