The answer is five factoral (5!) which is 120.
im assuming that any charcter can be a number or a letter: (24letters*10 possible numbers)^(4 digits)= 3317760000 possible combinations.
If the digits can repeat, then there are 256 possible combinations. If they can't repeat, then there are 24 possibilities.
3,124,550 possible combinations
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.
There are 210 4 digit combinations and 5040 different 4 digit codes.
128
There are infinitely many numbers and so infinitely many possible combinations.
140 possible combinations
In a 7 segment display, the symbols can be created using a selected number of segments where each segment is treated as a different element.When 1 segment is used, the possible positions are 7because it can be any of the 7 segments (7C1=7).When 2 segments are used, the number of possible combinations are 7C2=21.When 3 segments are used, the number of possible combinations are 7C3=35When 4 segments are used, the number of possible combinations are 7C4=35When 5 segments are used, the number of possible combinations are 7C5=21When 6 segments are used, the number of possible combinations are 7C6=7When 7 segments are used, the number of possible combinations are 7C7=1Adding the combinations, 7+21+35+21+7+1=127Therefore, 127 symbols can be made using a 7 segment display!
There are twelve possible solutions using the rule you stated.
im assuming that any charcter can be a number or a letter: (24letters*10 possible numbers)^(4 digits)= 3317760000 possible combinations.
If the digits can repeat, then there are 256 possible combinations. If they can't repeat, then there are 24 possibilities.
There are 5,040 combinations.
3,124,550 possible combinations
How many 3 diget number can made by the the following numbers 6789
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.
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.