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How many combinations of 9 digits are possible if no two adjacent digits are the same?
The 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.
What else can I help you with?
For her bicycle Lisa has a chain and a combination lock that has a 4-digit combination where each digit can range from 0 - 9 If adjacent digits must be different how many possible combinations can?
10x9x9x9
How many combinations of 5 are possible from 30 digits?
There are: 30C5 = 142,506
How many possible 4 digit combinations are possible using 4 digits?
It can be calculated as factorial 44! = 4x3x2x1= 60
What combinations can you make with numbers 1 4 7 2 5 8 0?
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.
How many possible combinations of three can be made from 100 digits?
100C3= 100x99x98/3!= 161700 The notation is 100 choose 3
For her bicycle Lisa has a chain and a combination lock that has a 4-digit combination where each digit can range from 0 - 9 If adjacent digits must be different how many possible combinations can?
10x9x9x9
How many different number combinations are in the world?
Since a number can have infinitely many digits, there are infinitely many possible combinations.
How many combinations of 5 are possible from 30 digits?
There are: 30C5 = 142,506
How many possible 4 number combinations can you get using numbers 0-9?
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.
How many sum of the digit is nine?
To determine how many numbers have a sum of digits equal to nine, we consider the possible combinations of digits that add up to nine. These combinations include (9,0), (8,1), (7,2), (6,3), and (5,4). Therefore, there are five possible combinations of digits that sum up to nine.
How many possible 4 digit combinations are possible using 4 digits?
It can be calculated as factorial 44! = 4x3x2x1= 60
What combinations can you make with numbers 1 4 7 2 5 8 0?
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.
How many possible 5 digit combinations are there using 5 digits?
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.
How many possible combinations is there with 1379?
To calculate the number of possible combinations of the digits 1, 3, 7, and 9, we can use the formula for permutations of a set of objects, which is n! / (n-r)!. In this case, there are 4 digits and we want to find all possible 4-digit combinations, so n=4 and r=4. Therefore, the number of possible combinations is 4! / (4-4)! = 4! / 0! = 4 x 3 x 2 x 1 = 24. So, there are 24 possible combinations using the digits 1, 3, 7, and 9.
How many 3 digit number are formed using the digits 3 5 7?
With repeating digits, there are 33 = 27 possible combinations.Without repeating any digits, there are 6 combinations:357375537573735753
How many possible unique alpha numeric combinations are there using all letters for 4 characters?
im assuming that any charcter can be a number or a letter: (24letters*10 possible numbers)^(4 digits)= 3317760000 possible combinations.
How many number combinations possible from a 13 digit number?
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.
