Since the smiling face is the one with a restriction, we could put that one in first. Since it can't go at either end, there are 3 options for its space:

 

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Then moving onto the next face, whichever of the 3 options we chose for the smiling face, there are 4 options for this face (any of the 4 remaining spaces).

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So since there are 4 options corresponding to each of the 3 options for the smiling face, that gives a total of 4$\times$3 = 12 options for the first two faces.

Then there will be 3 spaces left in which to put the next face, then 2 spaces, and finally only 1 space left for the last face.

So altogether there are 12$\times$3$\times$2$\times$1 = 72 options to place all of the faces.