---
title: "Systematic Listing"
description: "List all possible pairs made by choosing one option from each of two groups."
canonical: https://lightmysky.com/learn/mathematics/systematic-listing-mt_gu0NPDvlhY
source: https://lightmysky.com/learn/mathematics/systematic-listing-mt_gu0NPDvlhY.md
retrieved: 2026-11-12
---

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# Systematic Listing

List all possible pairs made by choosing one option from each of two groups.

Subject: Mathematics · Area: Algebra · Ages 10 to 11
Page: https://lightmysky.com/learn/mathematics/systematic-listing-mt_gu0NPDvlhY

## Ready when they can

- List all possible meal combinations from 3 starters and 4 mains
- Organise pairs in a table so each possible pair appears once
- Explain why the total equals the product of the two group sizes when every pairing is allowed

## Lesson: Listing every combination

You have used multiplication to find totals. Here, a combination means one choice from each of two groups. Two choices from the same group do not make a complete pair. When every pairing is allowed, multiply the two group sizes to count the different pairs. Adding the group sizes counts the separate options.

*(drawing: For 4 shirt colours and 4 trouser colours, each dot represents one outfit. Each shirt has 4 trouser choices: 4 x 4 = 16.)*

**Example.** A cafe offers 4 starters and 5 mains. Any starter can go with any main. Pair the first starter with all 5 mains. Repeat for each remaining starter. There are 4 rows of 5 pairs: 4 x 5 = 20 meals. Follow the same main order for every starter.

Label rows with hats and columns with scarves. In each cell, combine its row's hat and its column's scarf:
Hat  red scarf  blue scarf
cap  cap-red  cap-blue
hood  hood-red  hood-blue
For the hood row and blue scarf column, write hood-blue. Fill every cell once to include every pair. Finish one row before starting the next. Either group can go down the side. One fixed hat has 2 scarf pairs.

**Example.** Check the pairs as well as the entry count. With these hats and scarves, four entries could still repeat cap-red and leave out hood-blue. Compare each expected pair with the list. If all listed pairs are valid and different, subtract their count from the expected total to find how many are missing.

If a pair is unavailable, leave it out. For 4 ticket types and 6 days, there would be 4 x 6 = 24 pairs. If exactly one pair is unavailable, there are 24 - 1 = 23 available pairs. Adding one new day adds 4 pairs, one for each ticket type.

You can also work backwards from a total. Suppose every pattern comes in 4 colours and there are 12 different pairs. Group the pairs into rows of 4: 4, 8, 12. That makes 3 rows, so there are 3 patterns. This works because every pattern has all 4 colour choices.

**Recap.** Pair each option with every allowed option in the other group. Check that each pair appears once. When every pairing is allowed, multiply the two group sizes to check the total.

*Lesson, questions and flashcards written for this dataset.*

## Practice

16 questions on this page, each with its working shown.

## Optional connections

- [Long multiplication (age 10+)](https://lightmysky.com/learn/mathematics/long-multiplication-age-10-mt_XLP1IM3Qbb)

## Opens up

- [Counting Outcomes with the Product Rule](https://lightmysky.com/learn/mathematics/counting-outcomes-with-the-product-rule-mt_74gSbYIG8p)
