[RubikRevolution]

ok solving number 25 -
For those who didn't take it and don't know the question:
Call a set of integers spacy if it contains no more than one out of any three consecutive odd integers. How many subsets of{1,2,3,...,12}, including the empty set, are spacy?
(A) 121 (B) 123 (C) 125 (D) 127 (E) 129

First divide it into 5 cases: empty set, set of 1, 2, 3, 4; imposible for more than 4.
Empty set is easy - just 1.
set of 1; also easy enough - 12
for the rest this is like the "unfriendly customers" problem.
Instead of seats customers sit in, you have numbers being choosen.
With two numbers being choosen match one of the "customers" with two previous numbers so
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Where the bold numbers are selected and the underline represents the groups. As long as the first group never intersects with the second group, and they stay in that order, you can shift the underlined selection (with bold at the end) without having two numbers less than 2 apart.
This leaves us with 10 groups, 8 idenical not selected, and 2 which can't be swaped so: 10!/8!2! = 45
Likewise with three, there are now 8 groups, 6 idenical, 3 which can't be interchanged: 8!/3!5! = 56
and with 4: 6!/4!2! = 15
1+12+45+56+15=129
The answer is E.