Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively : in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats

Hi, I'll give it a try. Let me know if the answer matches with the OA. As 4 girls and 4 boys are to be seated alternatively, I assume there are 8 positions. (Generally, GMAT provides the positions data specifically). Approach: Ways in which John & Susan are Not Adjacent = Total ways (minus) Ways in which they're adjacent Part-1: We find total ways Total ways can be BGBGBGBG and GBGBGBGB 4 boys can sit in 4 places in 4! ways and 4 girls can sit in 4 places in 4! ways. This can be done in 4!*4! for each arrangement of BGBGBGBG and GBGBGBGB i.e. 2*4!*4!= 1152 .... (1) Part-2: Ways where John and Susan are together Consider them as a group. Remaining are 3 boys and 3 girls 3 boys and 3 girls can sit in 3!*3! ways John(J) and Susan(S) can be arranged in following ways - JS _ _ _ _ _ _ = 3!*3! ways _ JS _ _ _ _ _ = 3!*3! ways _ _ JS _ _ _ _ = 3!*3! ways _ _ _ JS _ _ _ = 3!*3! ways _ _ _ _ JS _ _ = 3!*3! ways _ _ _ _ _ JS _ = 3!*3! ways _ _ _ _ _ _ JS = 3!*3! ways Total= 7*3!*3! Similarly, for Susan-John combination another 7*3!*3! Combining above, total ways in which John & Susan are together= 2*7*3!*3! = 504 ..... (2) TOTAL WAYS IN WHICH JOHN & SUSAN ARE NOT ADJACENT = EQN (1) minus EQN (2) = 1152 - 504 = 648

Nicely done by adthedaddy. One note is that this is unlikely to be a GMAT problem because it is so involved. However, if you understand it, you should be able to tackle anything that comes your way.