Answer should be ${3^N}-{1}$.
Explanation:
The total possible combinations (i.e., a line may either be at fault (in $2$ ways i.e stuck at fault $0$ or $1$) or it may not be, so there are only $3$ possibilities for a line ) is ${3^N}$.
In only one combination the circuit will have all lines to be correct (i.e not at fault.)
Hence ${3^N}-{1}$ (as it has been said that circuit is said to have multiple stuck up fault if one or more line is at fault)