Counting Principles (AQA GCSE Further Maths)

Revision Note

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Product Rule for Counting

What is meant by counting principles?

  • Counting principles state that if there are m ways to do one thing and n ways to do another there are m × n ways to do both things
  • Applying counting principles allows us to ...
    • ... see patterns in real world situations
    • ... find the number of combinations or arrangements of a number of items
    • ... find the number of ways of choosing some items from a list of items
  • When you have a question like “How many ways…?” you should always look for the words “AND” and “OR”
    • “AND means ×” 
      “OR means +”  

 

How do I choose an item from a list of items AND another item from a different list of items?

  • If a question requires you to choose an item from one list AND an item from another list you should multiply the number of options in each list
    • In general if you see the word 'AND' you will most likely need to 'MULTIPLY'
  • For example if you are choosing a pen and a pencil from 4 pens and 5 pencils:
    • You can choose 1 item from 4 pens AND 1 item from 5 pencils
    • You will have 4 × 5 different options to choose from

 

How do I choose an item from a list of items OR another item from a different list of items?

  • If a question requires you to choose an item from one list OR an item from another list you should add the number of options in each list
    • In general if you see the word 'OR' you will most likely need to 'ADD'
  • For example if you are choosing a pen or a pencil from 4 pens and 5 pencils:
    • You can choose 1 item from 4 pens OR 1 item from 5 pencils
    • You will have 4 + 5 different options to choose from

How many ways can n different objects be arranged?

  • When considering how many ways you can arrange a number of different objects in a row it’s a good idea to think of how many of the objects can go in the first position, how many can go in the second and so on
  • For n equals 2 there are two options for the first position and then there will only be one object left to go in the second position so
    • To arrange the letters A and B we have
      • AB and BA
    • For n equals 3  there are three options for the first position and then there will be two objects for the second position and one left to go in the third position so
      • To arrange the letters A, B and C we have
        • ABC, ACB, BAC, BCA, CAB and CBA
      • For n objects there are n options for the first position, n minus 1 options for the second position and so on until there is only one object left to go in final position
      • The number of ways of arranging different objects is n cross times left parenthesis n minus 1 right parenthesis cross times left parenthesis n minus 2 right parenthesis cross times... cross times 2 cross times 1 
  • If the objects being arranged can be repeated in the list then we do not reduce each term by one
    • For example to arrange the letters A, B and C where each letter can be used more than once we have 
      • AAA, AAB, ABB, ...
      • There are three options for the first position, three options for the second and three for the third
      • There are 3 × 3 × 3 = 27 possibilities
    • For example to arrange any 4 out of the 26 letters of the English alphabet, where each letter can be used more than once
      • There are 26 options for the first position, 26 for the second, 26 for the third and 26 for the fourth
      • There are 26 × 26 × 26 × 26 = 264 possibilities

Exam Tip

  • Always read a question carefully and identify where it requires you to add or multiply before beginning the problem

Worked example

Work out how many ways a 4 digit PIN code be made using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9 if

i)
each digit can be used any number of times in the PIN code,
 
ii)
each digit can only be used once and the first digit cannot be 0, 1 or 9? 

 

i)

There are 10 different options for each position in the PIN code, and repeats are allowed, so the number of ways will be

10 cross times 10 cross times 10 cross times 10

bold 10 bold space bold 000

ii)

As the first digit cannot be 0, 1, or 9, there are 7 options for the first position in the code

The second position can be any number from 0 to 9, 10 options, but one of those numbers has been used already in the first position, so there are 9 options for the second position

As we are not allowed repeats, the number of options remaining reduces by 1 each time for the third and fourth positions

7 cross times 9 cross times 8 cross times 7

bold 3 bold space bold 528

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Amber

Author: Amber

Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.