Five boys and three girls, all different, are to sit in a row of 8 chairs. In how many ways can they sit so that no two girls sit next to each other?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 14400
To keep a group apart, first seat the others, then put the group into the gaps between them, at most one in each gap.
The 5 boys sit in a row in \(5!=120\) ways
5 boys leave 6 gaps: one before the first boy, 4 between boys, one after the last boy
The 3 girls take 3 of the 6 gaps, one girl in each, in order: \(^6P_3=6\times5\times4=120\) ways
Ways \(=120\times120=14400\)
Why the other options are incorrect
Option (1) 40320 — Incorrect. This is \(8!\), every seating, with girls allowed side by side.
Option (3) 2400 — Incorrect. This is \(5!\times\,^6C_3\), which chooses the gaps but does not arrange the 3 different girls in them.
Option (4) 2880 — Incorrect. This is \(5!\times\,^4P_3\); it uses only the 4 gaps between the boys, but the two end places can also take a girl, giving 6 gaps.