If $n>0$ ($n$ natural) and $A⊆B⊆C⊆\{1,2,3,...,n\}$, how many 3-tuples of sets $(A,B,C)$ there are?
For $n=1$ there are four of them and I think that for $n=2$ there are sixteen. I don't know how to figure out general formula, so I would appreciate some hint.
There are $4^n$ of them, because for each element there are four options, 1)goes in C and not in A or B 2)goes in B and C and not in A 3)goes in all A, B and C 4)Doesn't go in any of A, B or C. and there are n elements in all in the set {1,2,...}. Hope it helps:)