Momma’s Pizza Shed has the toppings ham, cheese, salami, chicken, mushrooms, tomato, bacon and pineapple.

Jessie wants 2 toppings on her pizza.

How many choices does she have?

At Pizza Place there are 100 toppings available.

How many choices does Jessie have here?

This problem challenges students to find the pattern that exists in the number of combinations of 2 toppings that can be made from a choice of 8 pizza toppings, 100 toppings and t toppings.

Some students may begin by first finding all possible ways of obtaining 2 toppings in the simpler case. However, they should be encouraged to find a more sophisticated way to solve the problem. The approach of counting by listing all possibilities, then counting more systematically is a theme that occurs across mathematics.

See also: Penny’s Pizza, Statistics, Level 4, Can Stack, Algebra, Level 5

### The Problem

Momma’s Pizza Shed has the toppings ham, cheese, salami, chicken, mushrooms, tomato, bacon and pineapple. Jessie wants 2 toppings on her pizza. How many choices does she have?

At Pizza Place there are 100 toppings available. How many choices does Jessie have here?

### Teaching Sequence

- Discuss the students' favourite pizza toppings.
- Talk about how you might choose various toppings if you are only allowed a restricted number.
*How many ways are there of choosing 2 toppings from 3 available?* - Pose the first part of the problem, and ask students to suggest possible approaches.
- Have the students work in groups on the problem, moving on to the 100 toppings’ problem as appropriate.
- Make the Extension case of t toppings available.
- Have students record solutions as they work.
- As students share their solutions, have them justify the approaches they have taken.

#### Extension to the problem

If a pizza shop has t toppings, in how many ways can Jessie choose 2 toppings?

### Solution

Start with the 8 toppings problem.

Ham could be chosen with cheese, salami, chicken, mushrooms, tomato, bacon and pineapple – a total of 7 pairs of toppings.

After having used up ham, cheese could be chosen with salami, chicken, mushrooms, tomato, bacon and pineapple – a total of 6 pairs of toppings.

After having used up ham and cheese, salami could be chosen with chicken, mushrooms, tomato, bacon and pineapple – a total of 5 pairs of toppings.

After having used up ham, cheese and salami- chicken could be chosen with mushrooms, tomato, bacon and pineapple – a total of 4 pairs of toppings.

After having used up ham, cheese, salami and chicken - mushrooms could be chosen with tomato, bacon and pineapple – a total of 3 pairs of toppings.

After having used up ham, cheese, salami, chicken and mushrooms - tomato could be chosen with bacon and pineapple – a total of 2 pairs of toppings.

After having used up ham, cheese, salami, chicken, mushrooms and tomato - bacon could be chosen with pineapple – a total of 1 pair of toppings.

So altogether we have 7 + 6 + 5 + 4 + 3 + 2 + 1 pairs of toppings. This can quickly be added by noting that 7 + 1 = 8, 6 + 2 = 8 and 5 + 3 = 8.

28 is the total.

In choosing 2 toppings from 100, the same strategy can be used. Add:

T = 99 + 98 + 97 + 96 + … + 3 + 2 + 1.

See also:

T = 99 + 98 + 97 + 96 + … + 3 + 2 + 1, and

T = 1 + 2 + 3 + … + 96 + 97 + 98 + 99.

Adding Ts together gives 99 lots of 100 (because 99 + 1 = 100, 98 + 2 = 100, etc.). So

2T = 99 x 100.

Hence T = 99 x 50 = 4950.

#### Solution to the extension

2 toppings from t available toppings can be solved:

T = (t – 1) + (t – 2) + … + 2 + 1, or

T = 1 + 2 + … + (t – 2) + (t – 1).

Adding gives 2T = (t – 1) x t.

Hence T = (t – 1)t/2.

Check this out for t = 8 and t = 100.