This question is related to the question found here.
A ferris wheel has five cars, each containing four seats in a row.
There are $20$ people ready for a ride. In how many ways can the ride
begin? What if a certain two people want to sit in different cars?
I did not solve it correctly. After searching for this problem in net, came across the following official solution in a Kansas State University page.
[reference: Question number 25 (page 2) of this Kansas State University page]
According the Kansas State University page, answers are
part (a): $20! \times 4!$
part (b): $18!\times 4!\times 20 \times 16$
This solution was convincing to me. But, the solution in careerbless.com is different and as given below
According the careerbless.com page, answers are
part (a): $\dfrac{20! \times 4!}{5!}$
part (b): $18!\times 4\times 16$
Could someone please help me to find the right approach and answer? Also help me to understand the reason why one answer went wrong?
Note: I have observed that the answers of the two sources differ by a factor $5!.$
To me, it appears that both the solutions assumes that the cars are identical and the order of arrangements in each car are important, but finally comes with different answers.
Best Answer
The KSU answer to (a) is wrong. The reason is that they invent (as intermediate step) "$5$ named groups" which are not really distinct, and fail to later anonymize these groups by dividing out the $5!$ repetitions of the groups.
Another way to the same result is to have all the seats on the Ferris Wheel numbered, and simply arrange the riders into the seats $20!$ ways. The net division by $5$ in the careerbless page arises from assuming that we aren't concerned about which car is at the bottom when the final passengers are loaded.
For part (b), numbering the seats again gives the result $20\cdot 16\cdot18!$ with the optional division by $5$ if the wheel position is not important (as in the careerbless result).