At The Movies
AO elaboration and other teaching resources
Work systematically to find all the possible combinations
This problem is first of all about being systematic. This can be done using a number of strategies. You might make an organized list, draw a picture or use equipment. The difficult thing for many students is to keep track of all the possibilities as they go along. This is especially difficult if they use equipment.
The maths behind the problem involves counting all the possible combinations.
John, Jo and Chris have got seats for the movies. In fact their seats are F5, F6, F7. In how many ways can they sit in those seats?
- 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?
- Sharing of solutions
- Focus on the methods that students have used to be systematic.
Extension to the problem
What would the answer be if another two had joined the three friends?
What if your whole class went?
What if the seats were in a circle?
There are 6 ways for the students to sit on the seats.
Here we give every student a chance to sit in F5. There are 3 choices for this seat. For each particular student in F5 we then have two choices for F6. Then we have only one choice left for F7. Note that 3 x 2 x 1 = 6 is the final answer.
Solution to the extension
The extension is possibly ambiguous. We meant to mean that five friends had five seats, though some students might take it to mean that the five students had just three seats. Either way is worth considering.
Here five friends can be seated in five seats in 5 x 4 x 3 x 2 x 1 = 120 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. You might like to let the class calculate it.
On the other hand if the five friends are to be seated in only three seats, then there are 5 ways of putting the first friend in a seat, four ways for the second and three for the third. So it looks as if there are 5 x 4 x 3 = 60 ways. For the 25 members of your class you get 25 x 24 x 23 = something much more manageable than the number in the last paragraph! But then only three get to see the film!