How many strings of length 12 can we compose using letters A, B, C, and D if every letter should appear at least once? How many strings of length 12 can we compose using letters A,B,C, and D if every letter
should appear at least once?

can someone walk me through this? I believe using the concept of the sieve formula is what i am supposed to use but I can't figure out where to start, my first thought is that 12! is the total number of permutations but then I know i have to take into account that each letter must be used once but I can't get any farther

EDIT: total number of strings with no restrictions is 4^12

please if someone can walk me through that would be great
Hint: you need the inclusion-exclusion principle.  How many total strings are there?  How many strings using only A, B, C are there?  It is easier to approach it this way, by removing the ones that have only three letters.  There are four times as many missing one letter, but the ones that have only A, B are overcounted in the subtraction.
I think the answer is 56(24 + 32).
since every letter should appear once so break the 12 letters into 8 + 4,
where 4 is compulsory (a combination of A,B,C,D so 4*3*2*1 = 24 arrangements).
And the rest 8 can be all A, all B , all C,all D so 4 choices for each position so 4*8 = 32.

So, total is 24 + 32 = 56.
Wat Say ??