pokemon steel weakness
Related Articles
Trending Articles
The 2 A’s must not be adjacent.
The 3 G’s must not be adjacent (i.e., no two G’s adjacent).
C’s are already placed non-adjacent, so they don’t violate.
A and G can be adjacent freely.
So after placing 3 non-adjacent C’s (20 ways), we have 5 remaining positions to place 2 A’s and 3 G’s, with **no two A’s adjacent** and **no two G’s adjacent**.
But wait: if we place 3 G’s in 5 positions with no two adjacent, is it even possible?
Try: placing 3 G’s with no two adjacent in 5 positions.
Let the G’s occupy 3 positions such that no two are adjacent.
Then gaps: between G’s must be at least one non-G.
Minimum length for 3 non-adjacent G’s: G _ G _ G → 5 positions. Only one such configuration up to shift: G G G with gaps — actually, only possible if they are in positions with separation.