Wiki User
∙ 14y agoThe first can be any one of 10 digits. For each of those . . .
The second can be any one of the nine digits not the same as the first. For each of those . . .
The third can be any one of the nine digits not the same as the second. For each of those . . .
The fourth can be any one of the nine digits not the same as the third. For each of those . . .
The fifth can be any one of the nine digits not the same as the fourth. For each of those . . .
The sixth can be any one of the nine digits not the same as the fifth. For each of those . . .
The seventh can be any one of the nine digits not the same as the sixth. For each of those . . .
The eighth can be any one of the nine digits not the same as the seventh. For each of those . . .
The ninth can be any one of the nine digits not the same as the eighth.
The total number of possibilities is: (10 x 9 x 9 x 9 x 9 x 9 x 9 x 9 x 9) = 430,467,210 .
If matching adjacent digits were allowed there would be 1,000,000,000 possibilities.
So the restriction has eliminated 56.95% of them.
Wiki User
∙ 14y ago10x9x9x9
There are: 30C5 = 142,506
It can be calculated as factorial 44! = 4x3x2x1= 60
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.
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.
10x9x9x9
Since a number can have infinitely many digits, there are infinitely many possible combinations.
There are: 30C5 = 142,506
0000-9999 (10x10x10x10 or 104) = 10,000 possible combinations allowing for repeated digits. If you are not able to repeat digits then it's 10 x 9 x 8 x 7 or 5,040 possible combinations without repeated digits.
It can be calculated as factorial 44! = 4x3x2x1= 60
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.
Possible 5 digit combinations using 5 digits only 1 time is 5! or 5*4*3*2*1 or 120. Using 5 digits where numbers can be used 5 times is 55 or 3125.
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.
With repeating digits, there are 33 = 27 possible combinations.Without repeating any digits, there are 6 combinations:357375537573735753
im assuming that any charcter can be a number or a letter: (24letters*10 possible numbers)^(4 digits)= 3317760000 possible combinations.
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.
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