Counting/Permutations

Junior Cycle Higher Level Mathematics revision notes with diagrams, key terms and self-check questions.

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Counting lets you find the total number of outcomes or arrangements without having to write down every possibility by hand. In the Junior Cycle exam, you will use multiplication rules, systematic tables, and tree diagrams to count outcomes and calculate probabilities.

The Fundamental Principle of Counting

The Fundamental Principle of Counting states that if there are mm first choices and nn second choices for each first choice, there are m×nm \times n outcomes.

This multiplication rule applies across any number of steps:

  • Multiply the choices at each step, provided each stage's count applies after every allowed earlier selection. Choosing between 33 shirts, 22 trousers, and 22 pairs of shoes gives 3×2×2=123 \times 2 \times 2 = 12 outfits.
  • When the number of options is the same at every step, use index form. Tossing a coin 55 times gives 25=322^5 = 32 possible outcomes.
  • Multiply when choices happen in sequence ('and'). Add counts for separate cases only when no outcome belongs to more than one case.

Systematic Sample Spaces

A sample space is the set of all possible outcomes of an experiment. Listing outcomes systematically prevents you from missing any.

Two main visual layouts help you organise outcomes:

  • A two-way table is ideal for combining two independent events, such as flipping a coin and spinning a red/blue spinner:
Coin \ SpinnerRed (RR)Blue (BB)
Heads (HH)(H,R)(H, R)(H,B)(H, B)
Tails (TT)(T,R)(T, R)(T,B)(T, B)
  • A tree diagram works best for successive stages. Each branch represents an option, and each path from root to leaf represents one complete outcome. For example, spinning Red or Blue followed by flipping a coin gives 2×2=42 \times 2 = 4 branch endpoints: (R,H)(R, H), (R,T)(R, T), (B,H)(B, H), and (B,T)(B, T).
A tree branches into Red and Blue, then Heads and Tails from each colour, ending in (R,H), (R,T), (B,H) and (B,T).
A tree branches into Red and Blue, then Heads and Tails from each colour, ending in (R,H), (R,T), (B,H) and (B,T).

Permutations and the Box Method

A permutation is an arrangement of items where the order matters. Drawing boxes for each position makes it easy to track your choices.

  • With repetition: options remain constant for every slot. A 44-digit passcode using digits 00 to 99 gives 10×10×10×10=104=10 00010 \times 10 \times 10 \times 10 = 10^4 = 10\,000 codes.
  • Without repetition: options decrease by 11 after each pick. Line up 55 runners across 55 lanes in 5×4×3×2×1=1205 \times 4 \times 3 \times 2 \times 1 = 120 ways.
  • Restricted positions: it often helps to count the restricted position first; check the remaining choices carefully. To form a 33-digit even number from {1,2,3,4,5}\{1, 2, 3, 4, 5\} without repeats, fill the units slot first (22 choices: 22 or 44), then the first slot (44 digits left), and the middle slot (33 digits left): 2×4×3=242 \times 4 \times 3 = 24 numbers.

Linking Counting to Probability

Counting gives you the exact values needed to work out theoretical probabilities. For equally likely outcomes:

P(Event)=Number of favourable outcomesTotal number of possible outcomesP(\text{Event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}
  • Use counting rules to find both the top and bottom values. For two dice, there are 6×6=366 \times 6 = 36 total outcomes. Getting two odd numbers gives 3×3=93 \times 3 = 9 outcomes, so P(both odd)=936=14P(\text{both odd}) = \frac{9}{36} = \frac{1}{4}.
  • Solve 'at least one' problems by subtracting from 11: P(at least one)=1−P(none)P(\text{at least one}) = 1 - P(\text{none}). Assuming results of different matches are independent (one result does not affect another match's probabilities), if a team has a 23\frac{2}{3} chance of not winning each match, the chance of no wins in 33 games is (23)3=827\left(\frac{2}{3}\right)^3 = \frac{8}{27}, so P(at least one win)=1−827=1927P(\text{at least one win}) = 1 - \frac{8}{27} = \frac{19}{27}.

Key terms

Fundamental Principle of Counting
If there are m first choices and n second choices for each first choice, there are m × n outcomes.
Sample Space
The complete set of all possible outcomes of an experiment.
Permutation
An ordered arrangement of a group of items where the sequence matters.
Two-Way Table
A grid display used to list all combined outcomes of two separate events.
Tree Diagram
A branching diagram that displays all possible outcomes across successive stages.
Outcome
A single possible result of an experiment.

Check yourself

  1. A lunch deal offers 4 sandwiches, 3 drinks, and 2 snacks. How many different meal combinations are possible?

    4×3×2=244 \times 3 \times 2 = 24 combinations.

  2. A fair coin is tossed 4 times. What is the total number of outcomes in the sample space?

    24=162^4 = 16 outcomes.

  3. How many 3-digit odd numbers can you form from {1, 2, 3, 4, 5} without repetition?

    The units place has 3 choices (1, 3, 5), leaving 4 then 3 choices: 3×4×3=363 \times 4 \times 3 = 36 numbers.

  4. A team plays 3 matches where win, draw, and lose are equally likely, and results of different matches are independent (one result does not affect another match's probabilities). What is the probability that they win none of the 3 matches?

    The chance of no win per match is 2/3, so P(no win)=(23)3=827P(\text{no win}) = \left(\frac{2}{3}\right)^3 = \frac{8}{27}.

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