Please suggest different ways to solve this question
Find the total number of possibilities of colouring:
2^9 = 512.
Subtract 2x2 red square possibilities - 4.
That gives 508.
Not sure how u get 417
This uses Principle of Inclusion-Exclusion. You will see that this will be similar to n(A \bigcup B \bigcup C \bigcup D)
There are four 2 \times 2 squares in the figure, name them A,B,C,D in clockwise order.
There are a total of 2^9=512 colourings possible. From these
Subtract: Case I: One of the squares is painted fully red. The remaining 5 squares can be filled in 2^5 ways. Thus we have 4\times 2^5 = 128 such colourings
Add: The above include Case II: Where two sets of squares are painted red. Here we have to distinguish adjacent pairs (A,B), (C,D), (A,D), (B,C) which gives us 4 \times 2^3 = 32 colourings, and diagonal pairs A,C and B,D which gives 2 \times 2^2=8 colurings.
Hence from this case we have a total of 40 colourings
Subtract: Case III: Three sets of squares are filled in red. This gives 4 \times 2 =8 colourings.
Add: Case IV: All four sets are filled in red: 1 colouring
Hence number of permissible colurings = 512-128+40-8+1=417
There will be too much overlapping of cases.