Eleven objects A , B , C , D , E , F , α , α , α , β and β are arranged in a row, then the probability that every β has two α as neighbors is

Step-by-step solution

Total cases = 11 ! 3 ! . 2 ! The number of favourable cases when β is placed in gap of α = 1 × 2 C 2 × 1 × 7 ! = 7 ! Required probability = 7 ! × 3 ! × 2 ! 11 ! = 1 660