Hard GCSE Systematic listing Questions

A coin has $2$ outcomes

Heads or Tails.

A dice has $6$ outcomes

The numbers $1$ to $6$.

List method: pair each coin with each dice

Keep the coin fixed and run through the dice.

Coin Heads

Six outcomes with Heads.

Count these

That is $6$ outcomes.

Coin Tails

Six outcomes with Tails.

Count these

Another $6$ outcomes.

Total from listing

So the list has $12$ outcomes.

Now the product rule

Multiply the coin outcomes by the dice outcomes.

Work it out

The product rule gives $12$.

Compare the two

Both methods give $12$.

Why they agree

Each of the $2$ coin results has $6$ dice partners.

That is exactly multiplication

Two rows of six is $2 \times 6$.

So the product rule is a shortcut for listing

The product rule counts the same outcomes without writing them all out.

State the conclusion

So listing and the product rule both give $12$ outcomes.

Each move is $1$ step or $2$ steps

We need the moves to add up to $4$ steps.

Organise by number of $2$-steps

Work systematically by how many $2$-steps are used.

No $2$-steps

Four single steps.

First way

One way uses all single steps.

One $2$-step and two $1$-steps

The $2$-step can be in different positions.

$2$-step first

Take the $2$ first.

$2$-step in the middle

Take the $2$ second.

$2$-step last

Take the $2$ last.

Two $2$-steps

Two $2$-steps make $4$.

That way

One way uses two $2$-steps.

Three $2$-steps?

That would overshoot $4$, so no.

Collect all the ways

These are all the routes to the top.

Count them

There are $5$ ways.

Check each reaches $4$

Every listed way adds to $4$ steps.

State the answer

So there are $5$ ways to climb the staircase.

Even numbers end in an even digit

The last digit must be even.

But the number cannot start with $0$

The $0$ restriction on the first digit interacts with the units, so split into cases.

Case $A$: the units digit is $0$

Handle units = $0$ separately.

Hundreds digit (not $0$, not used)

The $0$ is in the units, so the hundreds can be any of $1-9$: $9$ choices.

Tens digit

Two digits used ($0$ and the hundreds), so $8$ remain.

Case $A$ count

So Case $A$ gives $72$ numbers.

Case $B$: the units digit is $2$, $4$, $6$ or $8$

There are $4$ non-zero even digits for the units.

Hundreds digit

It cannot be $0$ and cannot equal the units, so $8$ choices remain from $1-9$.

Tens digit

Two digits used, so $8$ of the $10$ remain ($0$ now allowed here).

Case $B$ count

Units $\times$ hundreds $\times$ tens.

Work out Case $B$

So Case $B$ gives $256$ numbers.

Add the two cases

Total $=$ Case $A$ $+$ Case $B$.

Work it out

So there are $328$ numbers.

Why two cases

The digit $0$ behaves differently in the units and hundreds, so the split is needed.

State the answer

So there are $328$ three-digit even numbers with no repeated digit.

Every team plays every other once

Each match is a pair of teams.

Count from each team's view

Each team plays $5$ others, giving $6 \times 5 = 30$.

Each match involves two teams

The match $A$ vs $B$ is the same as $B$ vs $A$.

So each is counted twice

Divide to remove the double counting.

Divide

So there are $15$ matches.

Check by listing pairs

Label the teams $A$ to $F$.

From $A$

$5$ matches involving $A$.

From $B$

$4$ new ones.

From $C$

$3$ new ones.

From $D$

$2$ new ones.

From $E$

$1$ new one.

Add them

Add the counts.

Work it out

So there are $15$.

Both methods agree

It is the same as choosing $2$ teams from $6$.

State the answer

So $15$ matches are played.

Two common errors when listing

People often miss an outcome or write one twice.

A random order is risky

Jumping around makes both errors likely.

A system fixes a rule

A systematic order tells you exactly what comes next.

Example rules

For example, list numbers from smallest to largest.

Or fix one thing and vary another

Keep the first item fixed and change the second in order.

Because there is a clear next, nothing is skipped

Following the rule step by step means no outcome is left out.

Because you never go backwards, nothing repeats

Moving steadily forward means you never write the same one twice.

Example: two coins

Fix the first coin, then vary the second.

First coin $H$

List the second coin's options.

First coin $T$

Then move to the first coin being Tails.

All four appear once

The system guarantees each of the four appears exactly once.

A random attempt might fail

Listing randomly you might write $HT$ twice or forget $TH$.

So the system gives completeness

Every outcome is included.

And no repetition

Each outcome appears exactly once.

State the reason

So a systematic order avoids both missing outcomes and repeating them.