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
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.
2087
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.
To calculate the number of possible combinations for the number 24680 using each number only once, we can use the formula for permutations. There are 5 numbers to arrange, so the number of permutations is 5! (5 factorial), which is equal to 5 x 4 x 3 x 2 x 1 = 120. Therefore, there are 120 possible combinations for the number 24680 using each number only once.
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
To calculate the total number of possible combinations for a license plate using 3 letters and 3 numbers, we need to multiply the number of options for each character position. For letters, there are 26 options (A-Z), and for numbers, there are 10 options (0-9). Therefore, the total number of combinations can be calculated as 26 (letters) * 26 (letters) * 26 (letters) * 10 (numbers) * 10 (numbers) * 10 (numbers) = 17,576,000 possible combinations.
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.