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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.
If you wanted to know how many 5 digit combinations you could make using only 2 digits how would you solve this?
Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.
How many number combinations are possible with 0123?
16 if you include 0 starting 12 if not That is not correct. 0123 is a 4-digit number = (n) therefore n! = 4! = 1*2*3*4 = 24 (if 0 is included as I assume that is the question) 123 is a 3-digit number = (n) Therefore n! = 3!= 1*2*3 = 6
How many 4 digit combinations can be made using numbers 1 2 3 4 6?
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
How many license plates can be made using either three digits followed by three letters or three letters followed by three digits?
35,152,000 (assuming that 000 is a valid number, and that no letter combinations are disallowed for offensive connotations.) Also, no letters are disallowed because of possible confusion between letters and numbers eg 0 and O.
Using 4 digits what is the total number of combinations for a 4 digit lock code?
If the digits can repeat, then there are 256 possible combinations. If they can't repeat, then there are 24 possibilities.
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 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.
What are the possible different combinations of numbers in four digits?
If the numbers contain zeros, the total number of combinations is 10,000. You can work this out easily logically: For ten single-digit numbers (0,1,2,3,4,5,6,7,8,9) then there are 10 possible 'combinations' For numbers with 2 digits then for each possible digit in the 10s column (e.g. in the 20s range) there are another 10 possible combinations (20,21,22,23,24,25,26,27, 28,29). As there are 10 possible ranges (single digits, teens, twenties, thirties etc) there will be 10 X 10 or 100 possible combinations. using the same logic, for three digits, there will be 10 X 10 X 10 or 1000 digits. And for 4 digits there will be 10 x 10 x 10 x 10 = 10,000 possible combinations. So for a number, say, with x digits, the total number of combinations of those digits will be 10 x 10 x 10..... etc with x numbers of 10s in the calculation. You can find out the number of combinations of any set of letters or numbers in the same way. as an example, to find out, say, the possible combinations of letters in the alphabet of 26 letters, then using the same method this can be given as 26 x 26 x 26 x 26............. with 26 '26's' in a row multiplied together. This gives the staggering amount of approximately 615612 followed by 31 zeros.
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 4 digit combinations are possible using 4 digits?
It can be calculated as factorial 44! = 4x3x2x1= 60
How many 4 number combinations using digits 1357 one and only once?
There is only one combination. The order of the digits in combinations makes no difference. They are considered as being different if they are permutations, not combinations.
How many 9 number combinations using 0-9?
There are only 10 combinations. In each combination one of the 10 digits is left out.
How many 6 number combinations using digits 1 2 3 4 5 6?
654321-100000= 554321 combinations
What is the largest possible number using these digits 1 3579?
If the digits are only used once, each, the largest possible number would be 97,531
What are the 5-number combinations not using any number twice?
There are infinitely many numbers and so infinitely many possible combinations.
