7 Children are seated randomly in a row. What is the probability that Joe and Bob sit together
and Mimi and Lisa sit together, but not next to Joe and Bob?
Solution
total ways they can be seated = 7!
now if we consider Joe and Bob sit together and Mimi and Lisa sit together the we can consider
Joe and Bob as one child and Mimi and Lisa as one child
no of ways they ca be seated = 5!*2*2 // 2comes as they can sit in two differnt ways next to
each other
no of ways they sit next to other = 4!*2 // same logic as previous consider all four as one child
no of ways they dont seat nex to oher = 5!*2*2 -4!*2
probabilit = (5!*2*2 -4!*2 )/7! = .086.
Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...
7 Children are seated randomly in a row. What is the probability th.pdf
1. 7 Children are seated randomly in a row. What is the probability that Joe and Bob sit together
and Mimi and Lisa sit together, but not next to Joe and Bob?
Solution
total ways they can be seated = 7!
now if we consider Joe and Bob sit together and Mimi and Lisa sit together the we can consider
Joe and Bob as one child and Mimi and Lisa as one child
no of ways they ca be seated = 5!*2*2 // 2comes as they can sit in two differnt ways next to
each other
no of ways they sit next to other = 4!*2 // same logic as previous consider all four as one child
no of ways they dont seat nex to oher = 5!*2*2 -4!*2
probabilit = (5!*2*2 -4!*2 )/7! = .086
