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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? 1 0 comment Jay Patel
Assuming that cars are identical and order of arrangements inside a car matters.

Let's first assign $20$ persons into $5$ identical groups where each group must contain $4$ persons. Number of ways in which this can be done
$=\dfrac{20!}{\left(4!\right)^5×5!}$

The $5$ groups can be arranged around the circle in $4!$ ways.

In each group, $4$ persons can interchange their positions in $4!$ ways. Since we have $5$ groups, number of ways $=\left(4!\right)^5$

Therefore, number of ways the arrangements can be made without any restrictions
$=\dfrac{20!}{\left(4!\right)^5×5!}×4!×\left(4!\right)^5=\dfrac{20!×4!}{5!}$ [answer for part a]

Now let's examine the arrangements where two persons, A and B must sit in same car.

Select $2$ persons who will sit with A and B. This can be done in $18\text{C}2$ ways. These four persons constitute one group.

Remaining $16$ persons can be arranged into $4$ identical groups (with $4$ persons in each group) in $\dfrac{16!}{\left(4!\right)^4×4!}$ ways.

The $5$ groups can be arranged around the circle in $4!$ ways.

As we have seen earlier, in each group, $4$ persons can interchange their positions in $4!$ ways. Since we have $5$ groups, number of ways $=\left(4!\right)^5$

Therefore, number of ways the arrangements can be made where two particular persons must sit in same car
$=18\text{C}2×\dfrac{16!}{\left(4!\right)^4×4!}×4!×\left(4!\right)^5=18\text{C}2×16!×4!$

Hence, number of ways the arrangements can be made where two particular persons must not sit in same car
$\dfrac{20!×4!}{5!}-18\text{C}2×16!×4!\\ =\dfrac{18!×20×19}{5}-\dfrac{18×17}{2!}×16!×4!\\ =18!×4×19-18!×12\\ =18!×4×16$ 0 0 comment