N:Dure wrote:A ride at Disneyland can hold 2 people in the front seat and 3 in the back seat. In how many ways can 8 people be seated on the ride if Alice must sit in the left-front seat and Stephen must sit in the middle of the back seat?


Front seat 1 place left can be decided in 6 ways

Back seat 2 places 6 people: 6!/2! 4! = 15

15 * 6 = 90 ways

Is this correct?

No.
Once you've placed one person in the remaining front seat (right-front one), there are 5 people left. We have to place these 5 people in the remaining 2 back seats. These can be done in 5*4 = 20 ways. Because one of the 2 seats can be filled with any one of the 5 people and the remaining one seat can be filled with any one of the remaining 4 people.

Total number of ways = 6*5*4 = 120

Overall simple solution = Number of ways to fill up 3 seats by 6 people = (First one can be filled with any one of the 6)*(Second one can be filled with any one of the 5)*(Third one can be filled with any one of the 4) = 6*5*4 = 120

N:Dure wrote:Thanks Anurag. So for the 2nd part it's 5 people & 2 seats: 5!/3! = 20 (Why is this is a permutation? order doesn't matter here

The OA is also given as 90. Wrong answer?

Order does matter here.
The back seat is something like: _____ Stephen _____
Where the dashes are to be filled by two person say Jack and Jill.
Now the arrangement Jack Stephen Jill is not same as the arrangement Jill Stephen Jack. Thus order does matter.