Question: There are 6n flowers of one type and 3 flowers of another type, total no of garlands possible?
My approach-: I have considered 3 cases Case 1. Garland contains only 6n flowers No of possible ways $= \frac{(6n-1)!}{2}$
Case 2. Garland contains only type containing 3 flowers No of permutations$= \frac{2!}{2}$
Case 3. Garland has both types of flowers No of permutations $= \frac{(6n+2)!}{2*6n!*3!}$
Total permutations$= (6n+2)!/(2*6n!*3!) + 1 + (6n-1)!/2$
Is this approach correct or have I missed/considered some cases? What approach should be used for such questions?