32C3 = 4960
3 x 3 x 3 x 3 = 81 combinations
45
10,000.
5040, assuming none of the digits are the same. (Assuming they're not, there's 5040 unique combinations you can make out of 7 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.
32C3 = 4960
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
To calculate the number of 4-digit combinations that can be made with 4 digits, we can use the formula for permutations. Since there are 10 possible digits (0-9) for each of the 4 positions, the total number of combinations is 10^4, which equals 10,000. This is because each digit can be selected independently for each position, resulting in a total of 10 choices for each of the 4 positions.
3 x 3 x 3 x 3 = 81 combinations
45
Oh, dude, you're hitting me with some math vibes here. So, if you have 6 digits to choose from to make a 4-digit combination, you can calculate that by using the formula for permutations: 6P4, which equals 360. So, like, you can make 360 different 4-digit combinations from those 6 digits. Math is wild, man.
How many license plates can be made using either two uppercase English letters followed by four digits or two digits followed by four uppercase English letters?
10,000.
5040, assuming none of the digits are the same. (Assuming they're not, there's 5040 unique combinations you can make out of 7 digits).
9*9*9 = 729 using the digits 1 to 9 and 2*9 using 10 and another digit. 749 in all.
10,108,800 assuming that leading 0s are not allowed in the 3-digit numbers; that the licence authorities do not block some letter combinations because they are rude, and letters that might be confused with numbers (O and 0, I and 1, Z and 2, S and 5, B and 8) are all permitted.