To a certain conference, each firm can send two employee representatives, on the condition that one of them is a male and the other a female. If 15 firms were represented in this conference, what is the probability that no two females are seated next to each other? (Assume that all 30 members are seated in a single row of seats) A] 1/2 B] 14!/30! C] 15!/30! D] 28/30!
For above question I am trying following: 15 men can be arranged in 15! ways. There are 16 spots among men (-M-M-.....-M-). 16 spots can be taken by 15 women in 16! ways. My answer is (15!*16!)/30!. But it's not listed in answer choices. Am I missing something or answers are incorrect?
My approach is similar to approach in question in below link. How many ways are there for 10 women and six men to stand in a line