- #1

RChristenk

- 51

- 5

- 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.

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.