Yes. By definition 0!=10!=10!=1, which keeps nC0=1{}^{n}C_{0}=1nC0=1 and nCn=1{}^{n}C_{n}=1nCn=1 consistent, there is exactly one way to choose nothing, and one way to choose everything.

Permutation & Combination Practice

How to use these practice questions

The six questions below rise in difficulty across the whole chapter, from a single ordered choice to an $\text{``at least''}$ committee that needs several cases. Give yourself roughly four to seven minutes per question and work on paper first, writing every line the way you would in the real exam, first decide whether order matters, then quote ${}^{n}P_{r}$ or ${}^{n}C_{r}$ , substitute, and state the count.

Resist the urge to peek.

Only once you have committed to a full answer should you open the solution and mark yourself line by line. When your working differs from ours, stop at the exact line where the two part company; that single step is almost always where the mark was lost.

Because Add Math is marked analytically, a correct set-up still earns credit even when the arithmetic slips, so always write the permutation or combination you are computing before you reach for the calculator.

Six practice questions

Q1[2 marks]

A club has $7$ members. A president and a secretary are to be chosen, and no member may hold both posts.

In how many ways can this be done?

Show worked solution

The two posts are different, so order matters and this is a permutation. By the multiplication principle there are $7$ choices for president and then $6$ for secretary, which is ${}^{7}P_{2}$ :

{}^{7}P_{2}=\frac{7!}{(7-2)!}=\frac{7!}{5!}=7\times 6=42

Answer

There are $42$ ways. Check with the multiplication principle directly: $7$ ways to pick the president, $6$ ways left for the secretary, giving $7\times 6=42$ .

Q2[3 marks]

Find the number of different arrangements of all $5$ letters of the word $\text{PLANT}$ in a row. (All five letters are different.)

Show worked solution

Arranging all $n$ distinct objects in a row gives $n!$ arrangements.

With $5$ different letters, $n=5$ :

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

Answer

There are $120$ arrangements. Check by the multiplication principle: $5$ choices for the first letter, $4$ for the second, and so on, $5\times 4\times 3\times 2\times 1=120$ .

Q3[3 marks]

A committee of $3$ is to be chosen from $8$ students. In how many ways can the committee be formed?

Show worked solution

A committee has no ranked positions, so order does not matter and this is a combination, ${}^{n}C_{r}=\dfrac{n!}{r!\,(n-r)!}$ .

Substitute $n=8$ , $r=3$ :

{}^{8}C_{3}=\frac{8!}{3!\,5!}=\frac{8\times 7\times 6}{3\times 2\times 1}=\frac{336}{6}=56

Answer

There are $56$ committees. Check: ${}^{8}C_{3}={}^{8}C_{5}$ , and both equal $56$ , which is a useful symmetry check on the arithmetic.

Q4[4 marks]

Find the number of arrangements of all $6$ letters of the word $\text{NATURE}$ in a row such that the three vowels $(A, U, E)$ are always next to one another. (All six letters are different.)

Show worked solution

Keep the three vowels together by treating them as a single block. That block, together with the three consonants $N$ , $T$ , $R$ , makes $4$ units to arrange in a row:

4!=24 \quad\text{ways to arrange the units}

Within the block the three vowels can themselves be ordered in $3!$ ways.

Multiply the two counts:

4!\times 3!=24\times 6=144

Answer

There are $144$ arrangements. The key idea is that a $\text{``must be together''}$ restriction turns several objects into one block, then you re-arrange inside the block.

Q5[4 marks]

A team of $4$ is to be chosen from $5$ boys and $3$ girls. In how many ways can the team be chosen if it must contain exactly $2$ boys and $2$ girls?

Show worked solution

Choose the boys and the girls separately, since the two selections are independent, then multiply. Choose $2$ boys from $5$ and $2$ girls from $3$ :

{}^{5}C_{2}=\frac{5\times 4}{2\times 1}=10 \qquad {}^{3}C_{2}=\frac{3\times 2}{2\times 1}=3

{}^{5}C_{2}\times{}^{3}C_{2}=10\times 3=30

Answer

There are $30$ teams. Because a committee is unordered, each part is a combination, and independent choices are multiplied together by the multiplication principle.

Q6[5 marks]

A committee of $4$ is to be formed from $6$ men and $4$ women. Find the number of committees that contain at least $3$ men.

Show worked solution

$\text{``At least }3\text{ men''}$ out of a committee of $4$ means either exactly $3$ men (and $1$ woman) or exactly $4$ men (and $0$ women). Count each case, then add.

Exactly $3$ men:

{}^{6}C_{3}\times{}^{4}C_{1}=20\times 4=80

Exactly $4$ men (so no women are chosen, and ${}^{4}C_{0}=1$ ):

{}^{6}C_{4}\times{}^{4}C_{0}=15\times 1=15

The two cases cannot happen at once, so add them:

80+15=95

Answer

There are $95$ committees. Check the pieces: ${}^{6}C_{3}=20$ , ${}^{6}C_{4}=15$ , ${}^{4}C_{1}=4$ and ${}^{4}C_{0}=1$ ; adding the two cases gives $80+15=95$ .

How to mark yourself like an examiner

Add Math is marked analytically, which means marks are attached to steps, not only to the final number. When you check your own script, award yourself credit the way a marker would: look for the correct decision about order, the right set-up, and a clean final count.

Decision mark: did you correctly decide whether order matters, a permutation ${}^{n}P_{r}$ for ranked positions or arrangements, a combination ${}^{n}C_{r}$ for an unordered selection?
Set-up mark: is the expression written before any numbers are worked out, with the right $n$ and $r$ in place?
For a $\text{``together''}$ restriction: did you form a single block, count $k!$ for the units, and multiply by the internal arrangements?
For an $\text{``at least''}$ question: did you list every qualifying case and add them (or subtract the unwanted cases from the total)?
Answer mark: is the final whole number stated clearly? A count can never be a fraction, so any non-integer answer signals a slip to hunt down.

How a teacher helps

Marking yourself is powerful, but it is hard to see your own blind spots. In a one-to-one lesson our teacher watches the exact line where a mark slips away, a combination used where the posts are actually ranked, a $\text{``together''}$ block that forgot its internal $3!$ , or an $\text{``at least''}$ case quietly left out, and corrects the habit on the spot.

Because our teachers are experienced, you work with someone who explains why order matters in one question and not the next. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

Get 1-to-1 help.

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Frequently asked questions

How do I tell a permutation from a combination?

Ask whether order matters. If arranging in a line, or filling ranked posts such as president and secretary, order matters and you use ${}^{n}P_{r}$ .

If you are only choosing a group where no one has a special role, order does not matter and you use ${}^{n}C_{r}$ .

How do I handle letters that must stay together?

Glue them into a single block and count arrangements of the resulting units, then multiply by the number of ways to order the letters inside the block. For three letters kept together, that inside factor is $3!=6$ .

What does

\text{``at least''}

mean in a counting question?

It means every case from the stated minimum upwards. List each qualifying case, count it with combinations, and add the results.

When there are many cases, it is often faster to count the total and subtract the cases you do not want.

0!

really

1

Yes. By definition $0!=1$ , which keeps ${}^{n}C_{0}=1$ and ${}^{n}C_{n}=1$ consistent, there is exactly one way to choose nothing, and one way to choose everything.

Source:SRC-DSKP-EN

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

Permutation and Combination, Practice Questions

How to use these practice questions

Six practice questions

How to mark yourself like an examiner

How a teacher helps

Frequently asked questions

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