Permutations and Combinations Question

http://www.ocr.org.uk/Images/144608-question-paper-unit-4732-01-probability-and-statistics-1.pdf

Q 4c) and ii), answers are 24 and 54

For 4c, why can't I make each number 1(1) 1(2) 2(1) 2(2) 3(1) 3(2) and use 6x5x4/2!*2!*2! ?

It makes sense to me but it obviously doesn't give the answer, anyone have any good methods to deal with the weirder permutations and combinations questions? Thanks!

For c) The ways of including every number at most twice is the same as the total number of ways to make a number minus the ways in which all 3 digits are the same.
This gives the required answer in I think the easiest way.

How would I apply this in the second part or in a general example?

The number of all digits being the same or used three times is much more in ii