A railway carriage will accommodate 5 passengers on each side

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    Combinatorics
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Homework Help Overview

The discussion revolves around a combinatorial problem involving seating arrangements for passengers in a railway carriage, where specific seating preferences and restrictions are imposed on certain passengers. The subject area includes combinatorics and permutations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different methods for calculating the number of valid seating arrangements based on passenger restrictions. Questions arise regarding the independence of choices made for passengers facing forwards versus those facing backwards, and the implications of these choices on the overall calculation.

Discussion Status

Participants are actively engaging with the problem, offering alternative methods and questioning assumptions about the seating arrangements. Some have identified potential errors in calculations, while others are verifying their approaches. There is a productive exchange of ideas, though no consensus has been reached on a final solution.

Contextual Notes

Participants are working under the constraints of specific seating preferences for certain passengers, which complicates the arrangement calculations. There are indications of miscalculations that affect the outcomes being discussed.

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Homework Statement
A railway carriage will accommodate 5 passengers on each side (one side faces the engine, and one side faces backwards). In how many ways can 10 persons take their seats when two of them decline to face the engine, and a third cannot travel backwards?
Relevant Equations
Elementary combinatorics
The straightforward way to solve this question:

1. Two passengers decline to face the engine: ##^5C_2 \cdot 2! = 20##
2. One passenger cannot travel backwards: ##^5C_1 \cdot 1! =5##
3. The remaining seven passengers can sit in any order in the remaining seven seats: ## 7! ##

Multiply 1,2,3 together and get the correct answer 504000.

My question is what if I focused on the seven passengers that don't have any restrictions?
1. For the five seats facing backwards, from seven non-restricted passengers three can be chosen: ##^7C_3 \cdot 3! = 210##
2. For the five seats facing the engine, from seven non-restricted passengers four can be chosen: ##^7C_4 \cdot 4! = 840##
3. The two passengers that must face the back: ##2!##

Multiply 1,2,3 together and I get 352,800 which is wrong. Where did I make the mistake? Thanks.
 
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The three passengers facing backwards are not independent of the four passengers facing forwards: no passenger can do both! Also, a backwards facing seat could be occupied by one of those three undecided passengers, but also by one of the two who must face backwards. Thus, once the three undecided passengers who will face backwards are chosen, there are five people who can occupy any particular backward-facing seat. Your method fixes in advance which of the backward facing seats the three passengers (in some order) will sit in, with the other two taken (in some order) by the passengers who insist on facing backwards.

So: There are {}^5C_3 ways to select the backward-facing seats which the three chosen passengers will sit in. There are {}^7P_3 = 7!/4! = {}^7C_3 \cdot 3! ways in which three of the seven flexible passengers can occupy those seats. The other four passengers must face forwards; there are {}^5C_4 ways to choose the seats they sit in, and 4! ways for them to occupy the seats. Of the remaining passengers, the two who must face backwards can occupy the two remaining seats in 2! ways, and the one passenger who must face forwards takes the only remaining seat.

The number of permutations is then <br /> {}^5C_3 \cdot {}^7P_3 \cdot 2! \cdot {}^5C_4 \cdot 4! = \frac{5!}{3! \cdot 2!} \cdot \frac{7!}{4!} \cdot 2! \cdot \frac{5!}{4! \cdot 1!} \cdot 4! = 7! \cdot \frac{5!}{3!} \cdot \frac{5!}{4!} = 7! \cdot {}^5P_2 \cdot {}^5P_1.

It is easier to observe that choosing 4 of the 7 flexible passengers determines which way they all face, and then there are 5! ways for each group to order themselves, making {}^7C_4 \cdot (5!)^2 permutations,
 
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What you said in the end makes intuitive sense, but I calculated ##^7C_4 \cdot (5!)^2=3024000##, so why is the numerical answer wrong here?
 
RChristenk said:
What you said in the end makes intuitive sense, but I calculated ##^7C_4 \cdot (5!)^2=3024000##, so why is the numerical answer wrong here?
Check your calculation!
 
Ah it is correct indeed. I miscalculated ##^7C_4##.