With 3 digit numbers there are 10 digits (0, 1, ..., 9) from which to select 3 different digits. This can be done in:
10C3 = 10! / (10-3)!3! = 120 ways.
For a set with n items, r of them can be selected (permuted) in:
n x (n-1) x ... x (n - (r +1)) = n! / (n-r)! = nPr
ways. However, if the order of selection does not matter (just the combination of the r items) then the selected items can be chosen in
r x (r-1) x ... x 1 = r!
ways. Meaning there are
nPr / r! = nCr = n! / (n-r)!r!
combinations possible.
9!/6!, if the six different orders of any 3 digits are considered distinct combinations.
There are different numbers of combinations for groups of different sizes out of 9: 1 combination of 9 digits 9 combinations of 1 digit and of 8 digits 36 combinations of 2 digits and of 7 digits 84 combinations of 3 digits and of 6 digits 126 combinations of 4 digits and of 5 digits 255 combinations in all.
There are infinite combinations that can make 3879
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.
There are only five distinct odd digits.
There are 126 different 5 digit combinations. Note that the combination 12345 is the same as the combination 45312.
9!/6!, if the six different orders of any 3 digits are considered distinct combinations.
There are different numbers of combinations for groups of different sizes out of 9: 1 combination of 9 digits 9 combinations of 1 digit and of 8 digits 36 combinations of 2 digits and of 7 digits 84 combinations of 3 digits and of 6 digits 126 combinations of 4 digits and of 5 digits 255 combinations in all.
There are infinite combinations that can make 3879
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.
There are only five distinct odd digits.
1
The order of the digits in a combination does not matter. So 123 is the same as 132 or 312 etc. There are 10 combinations using just one of the digits (3 times). There are 90 combinations using 2 digits (1 once and 1 twice). There are 120 combinations using three different digit. 220 in all.
about 1,0000000000000
There is only one possible combination of a 13 digit number created from 13 digits. In a combination, the order of the digits does not matter so that 123 is the same as 132 or 312 etc. If there are 13 different digits (characters) there is 1 combination of 13 digits 13 combinations of 1 or of 12 digits 78 combinations of 2 or of 11 digits and so on There are 213 - 1 = 8191 in all. If the characters are not all different it is necessary to have more information.
For there to be palindromes, each digit must be replicated. Therefore there are at most three distinct digits.If there are 3 pairs of different digits, then there are 6 palindromes. If there can be more duplicate digits, then there are 27 palindromes.
If you have 4 positions, each of which can hold any of the ten digits, you have 10 to the power 4 combinations. If you can have only 4 different digits, you have 4 to the power 4 different combinations.