This is a tricky one, IMHO.
The number of numbers (I know, sounds crappy... ) made up with those four digits is 4^4 = 16^2 = 256 (four digits and four positions). Now we have to subtract the number of numbers (grrr.... ) that are greater than 4321. Well, first of all, 1 or 2 cannot be the hundreds digits.
Now suppose we're looking for numbers with the thousands digit 4 and the hundreds digit either 4 or 3 (meaning smth like 43xx and 44xx). The tens digit and the units digit can be either 1,2,3 or 4. For each of the two possible "formats" (43xx and 44xx) we have 2^4 (two spots to fill and four "candidates"), so this is in total 16*2 = 32 possible numbers. We must not however, forget about the numbers that have a 43xx format but are smaller or equal to 4321. These are: 4311, 4312, 4313, 4314, 4321 or 5 numbers. This means that from our initial 256 we need to subtract (32 - 5) = 27 numbers.
So our final result will be 256 - 27 = 229.
Took me 10 mins... I promise I will say screw it and guess it if this happens on the real thing!