smc

Since each of the elements of the subsets must be spaced at least two apart, a divider counting argument can be used. From the set $\{1,2,3,4,5,6,7,8,9,10,11,12\}$ we choose at most four numbers. Let those numbers be represented by balls. Between each of the balls there are at least two dividers. So for example, o | | o | | o | | o | | represents $1,4,7,10$. For subsets of size $k$ there must be $2(k-1)$ dividers between the balls, leaving $12-k-2(k-1)=12-3k+2$ dividers to be be placed in $k+1$ spots between the balls. The number of way this can be done is $\left(\genfrac{}{}{0ex}{}{(12-3k+2)+(k+1)-1}{k}\right)=\left(\genfrac{}{}{0ex}{}{12-2k+2}{k}\right)$. Therefore, the number of spacy subsets is $\left(\genfrac{}{}{0ex}{}{6}{4}\right)+\left(\genfrac{}{}{0ex}{}{8}{3}\right)+\left(\genfrac{}{}{0ex}{}{10}{2}\right)+\left(\genfrac{}{}{0ex}{}{12}{1}\right)+\left(\genfrac{}{}{0ex}{}{14}{0}\right)=\boxed{129}$.