Find the number of positive integers $$n$$ satisfying all the following conditions.
(a) The digits of $$n$$ lie in the set $$\{1, 2, 4, 8\}$$. (Digits may be repeated.)
(b) The sum of the digits is 14.
(c) If 1 occurs as a digit, it can occur only immediately to the right of 8.

At most one digit 8 can appear because $$8 + 8 > 14$$, so at most one digit 1 can appear, since every 1 needs its own preceding 8. But all permitted digits except 1 are even while the digit sum 14 is even, so no digit 1 occurs at all. With no 8, the counts $$r$$ of digit 2 and $$s$$ of digit 4 satisfy $$r + 2s = 7$$, giving $$1 + 6 + \binom{5}{2} + 4 = 21$$ numbers. With one 8 the rest sum to 6, so the digits are $$8, 2, 2, 2$$ with 4 arrangements or $$8, 2, 4$$ with $$3! = 6$$ arrangements, and the total is $$21 + 4 + 6 = 31$$.