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**. - DNSFLEX

Understanding the Context

Understanding the Constraints

You begin with 10 total character positions. Placing 3 non-adjacent C’s immediately restricts placement: no two C’s can be next to each other. This reduces your pool and forces careful spacing in the remaining 5 positions.

You must then fill these 5 spots with exactly:

• 2 A’s, and
  
• 3 G’s,
  while obeying:
  
• No two A’s adjacent (i.e., at least one character between any two A’s),
  
• No two G’s adjacent (same rule applied to G’s).

This dual adjacency restriction makes the problem non-trivial but solvable via structured enumeration and strategic placement.

Key Insights

Step 1: Place the 3 Non-Adjacent C’s

Denote the 10 positions as slots 1 through 10. Use a gap-based method to ensure no two C’s are adjacent:

• Imagine placing 3 C’s with at least one empty slot between each.
  
• This forces a minimum of 2 “spacer” slots between them.
  
• Total minimal slots used: 3 C’s + 2 spacers = 5.
  
• You now have 5 remaining positions for A and G.

The placement of the C’s defines critical gaps: positions between, before, and after the C’s become candidate zones.

Final Thoughts

Step 2: Allocate Positions Based on Available Slots

After placing 3 non-adjacent C’s, exactly 5 positions remain. Without loss of generality, define these by their spatial gaps relative to C’s. For example, if C’s are placed at positions 2, 5, 8:

• Available positions: 1, 3, 4, 6, 9, 10 (depending on exact C placement)
  
• The key is identifying connected runs where A’s and G’s can be distributed without violating adjacency rules.

20 Strategies to Guide Selection:

1. Map all possible triplets of 3 non-adjacent positions in 10 slots — ~82 combinations, but symmetry and constraints reduce this.
   
2. Group valid C-arrangements by gap sizes (e.g., isolated C’s, C’s with one spacer, C’s near ends).
   
3. For each arrangement, list the free 5 slots and classify their sequences (consecutive or fragmented).
   
4. Count total sequences where runs of ≤1 length occur for both A and G.
   
5. Use symmetry and reflection to avoid redundant enumeration.
   
6. Break down by the lengths of gaps between and around C’s — e.g., (0 gaps), (1 gap), (2+ spans).
   
7. Consider parity: odd vs even positions and how C placement breaks parity.
   
8. Apply recursive placement logic for A and G in remaining slots with adjacency checks.
   
9. Skip placements where any gap allows two A’s or two G’s once only two slots remain for multiple identical letters.
   
10. Use backtracking to prune invalid partial sequences early.
    
11. Leverage combinatorial formulas: C(n,k) avoiding adjacent repeats.
    
12. Group arrangements by relative spacing — e.g., C1-C2, C2-C3, and C1-C3 gaps.
    
13. Visualize placements as sliding windows or binary strings with constraints.
    
14. Apply inclusion-exclusion to subtract invalid keyframes.
    
15. Map placements onto integer partitions with separation rules.
    
16. Use dynamic programming over positions to track valid A/G counts and adjacency.
    
17. Validate each partial arrangement: scan for AA or GG blocks.
    
18. Once A and G counts are fixed (2 and 3), ensure no two same letters are adjacent.
    
19. Prioritize placements where A and G can “swap” roles without violating spacing.
    
20. Confirm all placements satisfy non-adjacency both for A and G simultaneously.