In how many ways can 6 green toys and 6 red toys be arranged, such that 2 particular red toys are never together whereas 2 particular green toys are always together?

Answer: Option D

Considering two green toys that are to be together as one unit.
We can arrange the 6 green toys and the remaining 4 red toys (excluding the 2 who are not to be together) is:
= 9! × 2! × ¹⁰C₂ × 2!
= 18 × 10!