262/2*102/2 = 67600/4 = 16900
32C3 = 4960
45
3 x 3 x 3 x 3 = 81 combinations
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.
45
3 x 3 x 3 x 3 = 81 combinations
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.
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.
45 In combinations, the order of the digits does not matter so that 12 and 21 are considered the same.