2006 AMC 12A Problem 25

By property $(2),$ a valid $k$-element set is a $k$-subset of $\{k, k+1, \ldots, 15\}$ with no two consecutive elements. Collapsing the gaps between chosen elements, these correspond bijectively to $k$-subsets of a $(17 - 2k)$-element set, counted by $\binom{17 - 2k}{k}.$ This is nonzero only for $k \le 5,$ so the total is $$\begin{gathered} \binom{15}{1} + \binom{13}{2} \\ {}+ \binom{11}{3} + \binom{9}{4} \\ {}+ \binom{7}{5} \\ = 15 + 78 + 165 + 126 + 21 \\ = 405. \end{gathered}$$ Thus, the correct answer is E.