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?

A.

11! × 2!

B.

9! × 90

C.

4 × 10!

D.

18 × 10!

• 9!*2!*10c2*2!
• 10 years agoHelpfull: Yes(1) No(0)
• 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:
  $= 9!×2!×{^10C_2}×2! =$ 18× 10!
• 9 years agoHelpfull: Yes(0) No(0)