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shivam nainwal

asked 2017-03-03 13:48:38

A roller coaster has five cars, each containing four seats, two in front and two in back. 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?

Jay Patel

answered 2017-03-06 18:40:11

Refer similar question for a ferris wheel where we deal with circular arrangements. This can be solved in the same way, but here we deal with linear arrangements and therefore it does not make any difference whether cars are distinct or not (since the positions of the are important, we are counting the distinct arrangements anyway)

Assuming that arrangements of the persons within the front seats matters. (similar assumption is taken for back seats as well)

Assign $20$ persons into $5$ identical groups where each group must contain $4$ persons. Number of ways in which this can be done

$=\dfrac{20!}{(4!)^5×5!}$

The $5$ groups can be arranged in $5!$ ways (as we are dealing with linear arrangements)

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

Therefore, number of ways the arrangements can be made without any restrictions

$=\dfrac{20!}{(4!)^5×5!}×5!×(4!)^5=20!$ [answer for part a]

This answer can be obtained directly, if we visualize that each different arrangement of the $20$ persons constitute one particular way of seating in the roller coaster, which is clearly $20!$

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 $\dfrac{16!}{(4!)^4×4!}$ ways.

The $5$ groups can be arranged in $5!$ ways (as we are dealing with linear arrangements).

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

Therefore, number of ways the arrangements can be made where two particular persons must sit in same car

$=18\text{C}2×\dfrac{16!}{(4!)^4×4!}×5!×(4!)^5=18\text{C}2×16!×5!$

Hence, number of ways the arrangements can be made where two particular persons must not sit in same car

$20!−18\text{C}2×16!×5!\\

=18!×20×19−\dfrac{18×17}{2!}×16!×5!\\

=18!×20×19−18!×60\\

=18!×20×16$

In fact, the second part can be solved easier. Seat A in any of the $20$ seats. Then B can be seated in any of the $16$ seats which is not near to A. Remaining $18$ persons can be arranged in $18!$ ways. Thus we get $20×16×18!$ directly.

Assuming that arrangements of the persons within the front seats matters. (similar assumption is taken for back seats as well)

Assign $20$ persons into $5$ identical groups where each group must contain $4$ persons. Number of ways in which this can be done

$=\dfrac{20!}{(4!)^5×5!}$

The $5$ groups can be arranged in $5!$ ways (as we are dealing with linear arrangements)

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

Therefore, number of ways the arrangements can be made without any restrictions

$=\dfrac{20!}{(4!)^5×5!}×5!×(4!)^5=20!$ [answer for part a]

This answer can be obtained directly, if we visualize that each different arrangement of the $20$ persons constitute one particular way of seating in the roller coaster, which is clearly $20!$

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 $\dfrac{16!}{(4!)^4×4!}$ ways.

The $5$ groups can be arranged in $5!$ ways (as we are dealing with linear arrangements).

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

Therefore, number of ways the arrangements can be made where two particular persons must sit in same car

$=18\text{C}2×\dfrac{16!}{(4!)^4×4!}×5!×(4!)^5=18\text{C}2×16!×5!$

Hence, number of ways the arrangements can be made where two particular persons must not sit in same car

$20!−18\text{C}2×16!×5!\\

=18!×20×19−\dfrac{18×17}{2!}×16!×5!\\

=18!×20×19−18!×60\\

=18!×20×16$

In fact, the second part can be solved easier. Seat A in any of the $20$ seats. Then B can be seated in any of the $16$ seats which is not near to A. Remaining $18$ persons can be arranged in $18!$ ways. Thus we get $20×16×18!$ directly.

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