John, Jo and Chris have seats for the movies.

In fact their seats are F5, F6, F7.

In how many ways can they sit in those seats?

Work systematically to find all the possible combinations

This problem requires a systematic approach to finding all possible combinations. Strategies might include making an organized list, drawing a picture or using equipment. Students should be encouraged to keep track of all the possibilities as they work.

### The Problem

John, Jo and Chris have seats for the movies. In fact their seats are F5, F6, F7. In how many ways can they sit in those seats?

### Teaching Sequence

- Use 3 students and 3 chairs to pose the problem.
- Ask the students to work in 3’s to solve the problem.
- Circulate to see that the students are keeping track of their solution.
*How are you recording your work?*

How do you know that you have found all the possible ways?

How could you convince someone else that you have found all the ways? - Share solutions
- Focus on the methods that students have used to be systematic.

#### Extension to the problem

How many ways are there if another two people join the three friends?

What if your whole class went?

What if the seats were in a circle?

### Solution

There are 6 ways for the students to sit on the seats.

F5 | F6 | F7 |

Chris | Jo | John |

Chris | John | Jo |

Jo | John | Chris |

Jo | Chris | John |

John | Jo | Chris |

John | Chris | Jo |

Each student has a chance to sit in F5. There are 3 choices for this seat. For each student in F5 there are two choices for F6. This leaves only one choice left for F7. Note that 3 x 2 x 1 = 6 is the final answer.

#### Solution to the extension

In the extension the number of seats is ambiguous. Both interpretations are worthy of investigation.

If five friends are to be seated in five seats there are 5 x 4 x 3 x 2 x 1 = 120 ways.

If five friends are to be seated in only three seats, there are 5 ways of putting the first friend in a seat, four ways for the second and three for the third. Therefore there are 5 x 4 x 3 = 60 ways.

If there are 25 in your class, then there are 25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1. This is a very big number. Your class might enjoy calculating this.