KBAT: Counting with Restrictions

Q: Can I earn marks if my final number is wrong?

Yes. Add Math Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. A correctly chosen block or gap method, with the units and internal orders shown, can score well even if an arithmetic slip changes the final total.

What makes this a KBAT question

A routine counting question asks 'how many arrangements of $n$ different items?' and the answer is $n!$ .

A KBAT question adds a condition, some items must be together, or no two of a group may be adjacent, and expects you to fold that condition into the count from the very first line, not to list arrangements and cross out the bad ones. That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: the multiplication rule, $n!$

and $^{n}P_{r}$ are ordinary Form 5 tools, but deciding to glue several items into a single block, or to seat one group first and slot the other into the gaps, is a modelling choice no one hands you. In Add Math this rewards students who can translate 'must be together' into a block and 'no two adjacent' into gaps, and who know that a well-chosen method removes the restriction instead of fighting it.

The choice of technique, not the arithmetic, carries the difficulty.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Try it yourself before reading the solution.

Q1[8 marks]

A photographer arranges $3$ adults and $4$ children, seven different people, in a single row for a photograph. Find the number of possible arrangements (a) with no restriction, (b) if the three adults must stand together, and (c) if no two adults may stand next to each other.

Show worked solution

Understand. All seven people are different, so order matters and every seat is distinct.

Part (a) is a plain arrangement; part (b) forces three people to be neighbours; part (c) forbids any two adults from being neighbours. We choose a method that builds each condition in from the start.

Plan. For (a) use $7!$ .

For (b) treat the three adults as one block, so there are $5$ units to arrange, and multiply by the internal orders of the block. For (c) seat the children first, then place the adults into the gaps between and around them so that no two adults share a gap.

Execute and check. (a) Seven different people in a row:

7!=5040

(b) Glue the three adults into a single block. That block, together with the $4$ children, makes $5$ units to arrange in a row, in $5!$

ways. Within the block the $3$ adults can be ordered in $3!$

ways:

5!\times 3!=120\times 6=720

Four children in a row create $5$ gaps, one before, three between, one after:

\_\;C\;\_\;C\;\_\;C\;\_\;C\;\_

Placing each adult in a different gap ensures no two adults are adjacent. Choose and fill $3$ of the $5$ gaps with the $3$ different adults in $^{5}P_{3}$ ways:

^{5}P_{3}=5\times 4\times 3=60

By the multiplication rule the number of arrangements with no two adults adjacent is:

4!\times{}^{5}P_{3}=24\times 60=1440

Check. The three answers should sit in a sensible order.

The 'together' count $720$ is small, as expected, because forcing three people to be neighbours is a heavy restriction. The 'no two adjacent' count $1440$ is larger but still well below the unrestricted $5040$ , which is right: forbidding adjacency removes many but not most arrangements.

Re-deriving $^{5}P_{3}=5!/2!=120/2=60$ confirms the gap step, and $24\times 60=1440$ is consistent.

Finding a sensible first step

When a counting question carries a restriction, the reliable first step is to name the condition and match it to a standard technique before you write any numbers. 'Must be together' points straight to the block method: tie the linked items into one unit, count the units, then multiply by the internal orders.

'No two adjacent' points to the gap method: seat the unrestricted group first, then drop the restricted items into the gaps so the condition is impossible to break. Deciding which technique fits is worth more than any factorial, because the right model makes the restriction disappear rather than something you police at the end.

Only after you have chosen the method do you count units, gaps and internal orders. Resist listing arrangements and subtracting, that route is slow and error-prone under exam pressure, and it earns no method mark for a clean idea.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a counting-with-restrictions question a marker looks for:

The unrestricted count set up correctly as $7!=5040$ .
The three adults treated as a single block for the 'together' case.
Both parts of the block count shown: $5!$ for the units and $3!$ for the order inside the block.
The gap method identified for 'no two adjacent', children seated first, gaps counted as $5$ .
The adults placed into the gaps as $^{5}P_{3}=60$ , not $^{5}C_{3}$ , because the adults are different.
The multiplication rule applied cleanly: $4!\times{}^{5}P_{3}=1440$ .

How a teacher helps

Counting improves fastest when a student can name the restriction and reach for the matching method without hesitating. In a one-to-one lesson our teachers ask you to say, in words, whether a group must stay together or must be kept apart, and to pick the block or gap method accordingly before touching a factorial.

We pay close attention to the choice between $^{n}P_{r}$ and $^{n}C_{r}$ , because whether order matters is where marks are most often lost. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

Message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

Why does the block method use

5!

and not

4!

Because after gluing the three adults into one block, that block is itself one unit to be arranged alongside the four children. That gives $4+1=5$ units in the row, arranged in $5!$

ways, and then $3!$ for the orders inside the block.

Counting only the four children would ignore the block's own position.

Why use

^{5}P_{3}

rather than

^{5}C_{3}

for the gaps?

Because the three adults are different people, so which adult goes into which gap matters. $^{5}C_{3}$ would only choose the three gaps; $^{5}P_{3}$ chooses the gaps and orders the distinct adults within them, which is what a row of a photograph needs.

Why seat the children first in the 'no two adjacent' part?

Because the restriction is on the adults, so we arrange the unrestricted group, the children, first. Four children create five gaps, and putting at most one adult in each gap makes it impossible for two adults to be neighbours.

Seating the restricted group first would leave you fighting the condition.

Can I earn marks if my final number is wrong?

Yes. Add Math Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line.

A correctly chosen block or gap method, with the units and internal orders shown, can score well even if an arithmetic slip changes the final total.

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026