Permutation & Combination Examples (Medium)

Q: When objects repeat, why do I divide instead of multiply?

Because identical objects make some of the n!n!n! arrangements indistinguishable. Swapping the 333 identical red discs produces the same row 3!3!3! times, so dividing by 3!3!3! removes the duplicates. In general the count is n!p! q!⋯\frac{n!}{p!\,q!\cdots}p!q!⋯n!.

Q: How do I decide between multiplying and dividing?

Independent stages that build one outcome are multiplied; identical objects that create duplicate arrangements are divided out. Reading the condition carefully tells you which situation you are in.

What these examples cover

These medium Permutation and Combination examples move past bare formulas to the conditions examiners actually attach: ‘these two must be together’, ‘some of the objects are identical’, and ‘the team must contain so many of each type’. The tools are the same multiplication principle, ${}^{n}P_{r}$ and ${}^{n}C_{r}$ from the easy set, now combined in two or three steps.

Numbers stay small and clean so the method, not the arithmetic, is what you practise. Cover each solution, attempt the question fully on paper, then check line by line; wherever your working parts from ours, isolate the exact step, because that is where the technique is really being tested.

Worked examples

Work through all three. Each one layers a condition onto a basic count, so read the question twice and decide what the condition changes before you reach for a formula.

Q1[3 marks]

In how many ways can $6$ different books be arranged in a row if two particular books must be placed next to each other?

Show worked solution

Tie the two particular books together and treat them as a single block. There are then $5$ items to arrange in a row, the block plus the other $4$ books, which can be done in $5!$

ways.

5! = 5 \times 4 \times 3 \times 2 \times 1 = 120

Inside the block, the two books can themselves be ordered in $2!$ ways.

By the multiplication principle, multiply the two counts:

5! \times 2! = 120 \times 2 = 240

Answer

There are $240$ arrangements. The block method works because tying the pair together keeps them adjacent; multiplying by $2!$

then restores the two internal orders we temporarily hid.

Q2[3 marks]

A row is made from $6$ coloured discs: $3$ are red, $2$ are blue and $1$ is green. Discs of the same colour are identical.

How many distinct arrangements are possible?

Show worked solution

If all six discs were different there would be $6!$ arrangements.

But the $3$ red discs are identical, so any arrangement is repeated $3!$ times by swapping the reds among themselves; likewise the $2$ blue discs repeat it $2!$

times. Divide the total by these repeats:

\frac{6!}{3!\,2!\,1!} = \frac{720}{6 \times 2 \times 1} = \frac{720}{12} = 60

Answer

There are $60$ distinct arrangements. Sanity check: with fewer identical discs there would be more distinct rows, and with all six identical there would be just one, dividing $720$ by $12$ lands sensibly between those extremes.

Q3[3 marks]

A committee of $4$ is to be chosen from $6$ boys and $4$ girls. In how many ways can it be formed if it must contain exactly $2$ boys and $2$ girls?

Show worked solution

The two requirements are independent choices, so count each and multiply. Choose $2$ boys from $6$ , order does not matter, so use a combination:

{}^{6}C_{2} = \frac{6!}{2!\,4!} = \frac{6 \times 5}{2 \times 1} = 15

Choose $2$ girls from $4$ the same way:

{}^{4}C_{2} = \frac{4!}{2!\,2!} = \frac{4 \times 3}{2 \times 1} = 6

By the multiplication principle, multiply the two selections:

{}^{6}C_{2} \times {}^{4}C_{2} = 15 \times 6 = 90

Answer

There are $90$ committees. Each choice of $2$ boys can be paired with each choice of $2$ girls, which is exactly what multiplying $15$ by $6$ counts.

Q4[3 marks]

In how many ways can $5$ students be seated in a row if two particular students refuse to sit next to each other?

Show worked solution

Count all unrestricted arrangements first, then subtract the arrangements where the two particular students do sit together. The unrestricted total is $5!= 120$ . For the ‘together’ count, tie the two students into one block, giving $4$ items to arrange, then multiply by the $2!$

orders inside the block:

4! \times 2! = 24 \times 2 = 48

Subtract the together count from the unrestricted total to leave only the arrangements that keep the two students apart:

120 - 48 = 72

Answer

There are $72$ arrangements. This is the mirror image of ‘must be together’: instead of tying the two students into a block, we remove the block count from the unrestricted total, since every arrangement is either together or apart.

Q5[4 marks]

Using the digits $1, 2, 3, 4, 5, 6$ without repetition, how many $4$ -digit odd numbers can be formed?

Show worked solution

A number is odd only if its units digit is odd, so fix that position first: the units digit must be one of $1, 3, 5$ , giving $3$ choices. The remaining three positions (thousands, hundreds, tens) are then filled, in order, from the $5$ digits left over:

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

By the multiplication principle, multiply the $3$ choices for the units digit by the $60$ ways to arrange the other three positions:

3 \times {}^{5}P_{3} = 3 \times 60 = 180

Answer

There are $180$ odd numbers. Fixing the restricted position (units) first, before arranging the free positions, is the standard move whenever a condition applies to only one place in the arrangement.

Q6[4 marks]

$6$ friends are to be seated at a round table. In how many ways can they be seated if two particular friends must sit next to each other?

Show worked solution

Tie the two particular friends into one block, leaving $5$ items to arrange around the table, the block plus the other $4$ friends. For $n$ items seated in a circle, rotations of the same seating are not counted as different, so the count is $(n-1)!$ :

(5-1)! = 4! = 24

Inside the block, the two friends can still be ordered in $2!$ ways.

Multiply the two counts:

4! \times 2! = 24 \times 2 = 48

Answer

There are $48$ seatings. The block method works the same way around a table as in a row, the only change is using $(n-1)!$

instead of $n!$ for the outer arrangement, because a circle has no fixed starting point.

Q7[3 marks]

A team of $4$ is to be chosen from $5$ boys and $3$ girls. In how many ways can the team be formed if it must contain at least $1$ girl?

Show worked solution

‘At least $1$ girl’ is easiest to count by complement: find the unrestricted total, then subtract the one case that has no girls at all. Choosing any $4$ from all $8$ students gives ${}^{8}C_{4} = 70$ .

The excluded case is choosing all $4$ members from the $5$ boys only:

{}^{5}C_{4} = \frac{5!}{4!\,1!} = 5

Subtract this all-boys case from the unrestricted total to leave only the teams with at least one girl:

70 - 5 = 65

Answer

There are $65$ teams. ‘At least one’ conditions are almost always faster by complement than by adding up every possible split ( $1$ girl, $2$ girls, $3$ girls) separately.

Q8[3 marks]

A debate team of $4$ students is chosen from $9$ students who tried out, and one of the $4$ is then appointed team captain. In how many ways can the team and captain be chosen?

Show worked solution

First choose the $4$ -member team from the $9$ students, order does not matter within the team, so this is a combination:

{}^{9}C_{4} = \frac{9!}{4!\,5!} = 126

Then choose which $1$ of those $4$ team members becomes captain; any of the $4$ chosen students can fill that role. Multiply the two counts:

126 \times 4 = 504

Answer

There are $504$ ways. Choosing the team is a combination because the $4$ members are unordered, but naming one of them captain adds a genuine choice on top, so that final step is multiplied in rather than folded into the ${}^{9}C_{4}$ .

Notice the shared habit across all three: name the condition, decide whether it splits the problem into stages (multiply) or hides repeats (divide), and only then substitute. Conditions do not change which formula exists, they change how many times you use it.

Key method points

These three examples rehearse the conditions that turn a plain count into a typical exam question. Keep the following points in mind as you practise more.

‘Must be together’, tie the objects into one block, arrange the blocks, then multiply by the internal arrangements of the block.
Arranging $n$ objects where some are identical gives $\frac{n!}{p!\,q!\cdots}$ ; divide by the factorial of each group of identical objects.
‘Choose so many of each type’, count each independent selection with ${}^{n}C_{r}$ and multiply the results.
Combinations are used when the order within a chosen group does not matter, which is the usual case for committees and teams.
Break a conditioned problem into clear stages; the multiplication principle links the stages together.
Because marking is analytic, each correct selection or arrangement line can earn method marks on the way to the final total.

How a teacher helps

The medium questions trip students not on the formula but on the condition, forgetting to multiply the block by $2!$ , or dividing by the wrong factorial when objects repeat. In a one-to-one lesson our teacher slows the moment where the condition is read, and has the student state in words what it changes before any number is written.

Because our teachers are experienced, you get someone who has seen these exact slips many times and knows how to fix the habit. Lessons are taught in English, while SPM papers are set in both Malay and English, so the working transfers cleanly to either paper.

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Frequently asked questions

Why do I multiply by

2!

in the block method?

Tying two objects into a block forces them to be adjacent, but it temporarily hides the two ways they can be ordered inside the block. Multiplying the block arrangements by $2!$

puts those two internal orders back, giving the correct total.

When objects repeat, why do I divide instead of multiply?

Because identical objects make some of the $n!$ arrangements indistinguishable.

Swapping the $3$ identical red discs produces the same row $3!$ times, so dividing by $3!$

removes the duplicates. In general the count is $\frac{n!}{p!\,q!\cdots}$ .

In a ‘so many of each type’ problem, why do I multiply the two combinations?

Because the choices are independent: every way of choosing the boys can be paired with every way of choosing the girls. The multiplication principle then gives ${}^{6}C_{2} \times {}^{4}C_{2} = 90$ .

How do I decide between multiplying and dividing?

Independent stages that build one outcome are multiplied; identical objects that create duplicate arrangements are divided out. Reading the condition carefully tells you which situation you are in.

Source:SRC-DSKP-EN

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

Permutation and Combination, Worked Examples (medium)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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