(a) In how many ways can 6 people be lined up to get on a bus?
Answer = 6!
(b) If 3 specific persons, among 6, insist on following each other, how many ways are possible?
Answer = 4! * 3!
(c) If 2 specific persons, among 6, refuse to follow each other, how many ways are possible ?
Answer = 6! – 5! * 2
I can't figure out how we got the answer of part (c). May you explain how to think in order to solve such question
Best Answer
Subtract the sitting together ways from the total possible ways. Let $T = 6$ people be lined up to get on a bus is then the possible ways $= 6!$ $R = 2$ person want to stick together then the possible ways $= 2*5!$. If $2$ specific persons, among $6$, refuse to follow each other,then the possible ways are
$= T - R = 6! - 2*5!\ \ \ \ $