Permutation & Combination Examples (Easy)

Q: Why is {}^{n}C_{r} smaller than {}^{n}P_{r}?

Because each group of rrr objects can be ordered in r!r!r! ways, all of which count as separate permutations but as a single combination. So nCr=nPrr!{}^{n}C_{r} = \frac{{}^{n}P_{r}}{r!}nCr=r!nPr, which is smaller for r≥2r \ge 2r≥2.

What these examples cover

These easy Permutation and Combination examples build the four moves that open the whole chapter: counting outcomes stage by stage with the multiplication principle, arranging a full set of distinct objects in a row using factorials, arranging only some of them where the order matters with ${}^{n}P_{r}$ , and choosing a group where the order does not matter with ${}^{n}C_{r}$ . Every question uses small, clean numbers so you can follow each line without leaning on a calculator.

Use the set the honest way: cover the solution, attempt the question fully on paper, and only then check line by line. Where your working differs from ours, find the exact step that parted, that single line is usually where the real learning sits.

Worked examples

Work through all four. Attempt each fully before you read the matching solution, and keep asking the one question that decides everything in this chapter: does the order of the objects matter?

If it does, you are permuting; if it does not, you are combining.

Q1[2 marks]

A sandwich shop lets you build one sandwich by choosing one of $4$ breads, one of $3$ fillings, and one of $2$ sauces. How many different sandwiches are possible?

Show worked solution

The sandwich is built in three independent stages, and the choice at one stage does not change the choices at the others. The multiplication principle says the total number of outcomes is the product of the number of choices at each stage.

4 \times 3 \times 2 = 24

Answer

There are $24$ different sandwiches. Check the idea in stages: each of the $4$ breads pairs with $3$ fillings to give $12$ bread–filling pairs, and each of those pairs with $2$ sauces, giving $12 \times 2 = 24$ .

Q2[2 marks]

In how many ways can $6$ different medals be arranged in a row on a display shelf?

Show worked solution

All six objects are different and every one is used, so this is a full arrangement of $6$ distinct objects. The first position can be filled $6$ ways, the next $5$ ways, and so on down to $1$ ; the total is $6!$

(six factorial).

6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720

Answer

There are $720$ arrangements. Notice the pattern: filling positions one at a time removes one medal each time, so the counts step down $6, 5, 4, 3, 2, 1$ and their product is $720$ .

Q3[3 marks]

From $6$ different books, in how many ways can $3$ of them be arranged in a row on a shelf?

Show worked solution

Here order matters, a different arrangement of the same three books counts separately, but only $3$ of the $6$ books are used. This is a permutation of $3$ objects chosen from $6$ , written ${}^{6}P_{3}$ .

Given in the exam

{}^{n}P_{r} = \frac{n!}{(n-r)!}

Substitute $n=6$ and $r=3$ :

{}^{6}P_{3} = \frac{6!}{(6-3)!} = \frac{6!}{3!} = \frac{6 \times 5 \times 4 \times 3!}{3!} = 6 \times 5 \times 4 = 120

Answer

There are $120$ arrangements. A quick check: fill the three shelf positions directly, $6$ choices, then $5$ , then $4$ , giving $6 \times 5 \times 4 = 120$ .

Q4[3 marks]

From a class of $7$ students, a committee of $3$ is to be chosen. In how many ways can the committee be formed?

Show worked solution

A committee is just a group, naming the same three students in a different order does not create a new committee, so order does not matter. This is a combination of $3$ chosen from $7$ , written ${}^{7}C_{3}$ .

Given in the exam

{}^{n}C_{r} = \frac{n!}{r!\,(n-r)!}

Substitute $n=7$ and $r=3$ :

{}^{7}C_{3} = \frac{7!}{3!\,(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35

Answer

There are $35$ committees. Check against the ordered count: arranging $3$ from $7$ gives ${}^{7}P_{3} = 7 \times 6 \times 5 = 210$ , and each committee has been counted $3!= 6$ times, so $210 \div 6 = 35$ .

Q5[2 marks]

A canteen offers $3$ types of rice set meals or $4$ types of noodle set meals for lunch. A student chooses exactly one set meal, either a rice set or a noodle set.

In how many ways can the student choose a set meal?

Show worked solution

The two menus are mutually exclusive alternatives, the student picks one set meal from one menu or the other, never both, so this is the addition principle, not the multiplication principle. Add the number of choices from each menu.

3 + 4 = 7

Answer

There are $7$ ways to choose a set meal. Contrast this with a multiplication situation: if the student were choosing a rice set AND a drink, the choices would multiply; here one dish is chosen from two separate menus, so the choices add.

Q6[3 marks]

Using all of the digits $2, 2, 3$ and $5$ , how many different $4$ -digit numbers can be formed?

Show worked solution

If the four digits were all different, they could be arranged in $4!$ ways.

But the digit $2$ appears twice, so swapping the two $2$ 's gives the exact same number, every arrangement has been counted twice.

4! = 24

Divide by $2!$ to remove the repeated count caused by the two identical $2$ 's:

\frac{4!}{2!} = \frac{24}{2} = 12

Answer

There are $12$ different $4$ -digit numbers. None of the digits is $0$ , so every one of the $12$ arrangements is a valid $4$ -digit number, no arrangement needs to be discarded.

Q7[3 marks]

A box contains $4$ different red sweets and $3$ different green sweets. In how many ways can a selection of $2$ red sweets and $1$ green sweet be made?

Show worked solution

Within each colour, a selection is just a group, order does not matter, so use combinations for each colour separately. Choosing the red sweets and choosing the green sweet are independent stages, so multiply the two counts (multiplication principle).

{}^{4}C_{2} \times {}^{3}C_{1}

Evaluate each combination, then multiply:

\frac{4!}{2!\,2!} \times \frac{3!}{1!\,2!} = 6 \times 3 = 18

Answer

There are $18$ ways to make the selection. Check the first factor alone: choosing $2$ sweets from $4$ gives ${}^{4}C_{2}=6$ red-sweet pairs, and each pair can go with any of the $3$ green sweets, giving $6 \times 3 = 18$ .

Q8[2 marks]

Without using a calculator, evaluate $\dfrac{8!}{6!}$ .

Show worked solution

Writing $8!$ as $8 \times 7 \times 6!$

lets the common $6!$ factor cancel with the $6!$

in the denominator, so neither factorial needs to be computed in full.

\frac{8!}{6!} = \frac{8 \times 7 \times 6!}{6!} = 8 \times 7

8 \times 7 = 56

Answer

$\dfrac{8!}{6!} = 56$ . This is the same pattern as ${}^{8}P_{2} = \frac{8!}{(8-2)!} = \frac{8!}{6!}$ , so the value also equals the number of ways to arrange $2$ objects chosen from $8$ .

See the single decision that separates these four: after the multiplication principle sets up the counting, the only question is whether order matters. Order matters for ${}^{n}P_{r}$ and for full arrangements; it does not for ${}^{n}C_{r}$ .

Get that decision right and most easy marks in this chapter follow.

Key method points

These four examples rehearse the everyday skills that open almost every Permutation and Combination question in Add Math. Keep the following points in mind as you practise more.

Use the multiplication principle when a task is done in independent stages: multiply the number of choices at each stage.
Arranging all $n$ distinct objects in a row gives $n!$ arrangements.
When order matters and you use $r$ of $n$ objects, use ${}^{n}P_{r} = \frac{n!}{(n-r)!}$ .
When order does not matter, use ${}^{n}C_{r} = \frac{n!}{r!\,(n-r)!}$ ; every combination is counted $r!$ times among the permutations.
Decide ‘does order matter?’ before choosing a formula, that one question sorts permutation from combination.
Because marking is analytic, a correct formula line with a clear substitution can earn method marks even if the final arithmetic slips.

How a teacher helps

When a student loses a mark here, it is almost always the order question answered the wrong way, reaching for ${}^{n}C_{r}$ on a seating problem, or ${}^{n}P_{r}$ on a committee. In a one-to-one lesson our teacher pauses at that exact decision and has the student say out loud whether order matters before any formula is written.

Because our teachers are experienced, you work with someone who explains the reasoning, not just the answer. Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation reads the same to you either way.

Get 1-to-1 help.

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Frequently asked questions

How do I know whether to use permutation or combination?

Ask whether the order of the chosen objects matters. If rearranging the same objects makes a genuinely different outcome, like seats in a row or positions in a race, use permutation ${}^{n}P_{r}$ .

If the order makes no difference, like members of a committee, use combination ${}^{n}C_{r}$ .

What does the multiplication principle actually say?

If a task is carried out in stages and the stages are independent, the total number of ways is the product of the number of choices at each stage. Four breads, three fillings and two sauces give $4 \times 3 \times 2 = 24$ sandwiches.

Why is

{}^{n}C_{r}

smaller than

{}^{n}P_{r}

Because each group of $r$ objects can be ordered in $r!$ ways, all of which count as separate permutations but as a single combination.

So ${}^{n}C_{r} = \frac{{}^{n}P_{r}}{r!}$ , which is smaller for $r \ge 2$ .

Are the

{}^{n}P_{r}

and

{}^{n}C_{r}

formulas given in the exam?

Both formulas appear on the SPM formula list, so you will not need to memorise them from scratch. Even so, knowing them well helps you choose the right one quickly and substitute without hesitation.

Source:SRC-DSKP-EN

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

Permutation and Combination, Worked Examples (easy)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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