Permutation & Combination Examples (KBAT)

Q: When do I add cases and when do I multiply?

Add when the cases are alternatives that cannot happen together, such as ‘exactly 3 women’ or ‘exactly 4 women’. Multiply when independent stages combine to build one outcome, such as choosing the men and choosing the women for the same team.

What these examples cover

These harder Permutation and Combination examples ask you to hold two ideas at the same time. The first forms even numbers from a set of digits, so the units digit is fixed first and the rest arranged around it.

The second counts committees that satisfy an ‘at least’ condition, which splits into separate cases that are added, not multiplied. The third selects a group with a constraint and then arranges the chosen members, joining combination and permutation in one problem.

Nothing here needs a new formula, only careful order of thinking. Attempt each fully on paper, then check line by line and pin down the exact step wherever your route differs from ours.

Worked examples

Work through all three slowly. The skill being tested is sequencing: which condition to satisfy first, and whether the separate parts should be added or multiplied.

Q1[4 marks]

How many $3$ -digit even numbers can be formed from the digits $1, 2, 3, 4, 5, 6$ if no digit may be repeated?

Show worked solution

A number is even when its units digit is even, so deal with that most-restricted position first. The units digit must be one of $2, 4, 6$ , giving $3$ choices.

\text{units digit: } 3 \text{ choices}

After the units digit is fixed, $5$ digits remain for the other two positions, and order matters. Fill the hundreds place ( $5$ choices), then the tens place ( $4$ choices):

\text{hundreds and tens: } 5 \times 4 = 20

By the multiplication principle, multiply the choices for the three positions:

3 \times 5 \times 4 = 60

Answer

There are $60$ such numbers. Check by symmetry: there are ${}^{6}P_{3} = 6 \times 5 \times 4 = 120$ three-digit numbers with no repeated digit, and exactly half end in an even digit because $3$ of the $6$ digits are even, so $120 \div 2 = 60$ .

Q2[4 marks]

A team of $4$ is chosen from $5$ men and $4$ women. In how many ways can the team be formed if it must contain at least $3$ women?

Show worked solution

‘At least $3$ women’ means the team has either exactly $3$ women or exactly $4$ women. These cases cannot happen together, so count each and add.

First, exactly $3$ women and $1$ man:

{}^{4}C_{3} \times {}^{5}C_{1} = 4 \times 5 = 20

Next, exactly $4$ women and $0$ men (the whole team is women):

{}^{4}C_{4} \times {}^{5}C_{0} = 1 \times 1 = 1

The cases are mutually exclusive, so add them:

20 + 1 = 21

Answer

There are $21$ teams. The key move is recognising that ‘at least’ opens two separate cases: adding covers ‘this case or that case’, while multiplying earlier covered ‘this choice and that choice’.

Q3[4 marks]

From $5$ senior members and $3$ junior members, a panel of $3$ is chosen and then seated in a row of $3$ chairs. The panel must contain exactly $2$ seniors and $1$ junior.

In how many ways can this be done?

Show worked solution

Split the task into two stages: first choose the panel (order does not matter), then seat the chosen three (order matters). Choose $2$ seniors from $5$ and $1$ junior from $3$ :

{}^{5}C_{2} \times {}^{3}C_{1} = \frac{5 \times 4}{2 \times 1} \times 3 = 10 \times 3 = 30

For each of these $30$ panels, the $3$ chosen members can be seated in a row in $3!$ ways:

3! = 3 \times 2 \times 1 = 6

By the multiplication principle, multiply the selection count by the arrangement count:

30 \times 6 = 180

Answer

There are $180$ ways. The two-stage structure is the point: combination first to decide who is on the panel, then permutation to decide the order they sit, selecting and arranging are different questions and each needs its own count.

Q4[5 marks]

Six committee members are to be seated around a round table for a meeting. The chairperson and the secretary insist on sitting next to each other.

In how many ways can the six members be seated?

Show worked solution

Because the chairperson and secretary must sit together, glue them into a single combined unit first. Together with the other 4 members, this gives 5 units to arrange around the table.

\text{5 units around a round table: } (5-1)! = 4! = 24

This treats the glued pair as having one fixed internal order. The chairperson and secretary can still swap places inside their unit, so multiply by the number of orders within the unit:

2! = 2

By the multiplication principle, combine the circular arrangement of the units with the internal swap:

24 \times 2 = 48

Answer

There are 48 seating arrangements. Check: with 6 distinct people around a round table there are $(6-1)!= 120$ arrangements in total, and two chosen people sit next to each other in exactly $\frac{2}{6-1} = \frac{2}{5}$ of them, giving $120 \times \frac{2}{5} = 48$ \u2014 the same answer.

Q5[4 marks]

A code is formed by arranging all 7 characters P, P, P, P, Q, Q, R in a row \u2014 that is, 4 identical letters P, 2 identical letters Q, and 1 letter R. How many different codes can be formed if the code must begin with the letter R?

Show worked solution

The 7 characters are not all different, so this is a permutation with repeated objects. Deal with the condition first: since there is only 1 letter R, fixing it at the front of the row can be done in exactly 1 way, leaving the remaining 6 positions to be filled by the 4 P's and 2 Q's.

\text{position 1: fixed as R, 1 way}

Arrange the remaining 4 P's and 2 Q's in the other 6 positions. Because the P's are identical and the Q's are identical, divide the ordinary $6!$

by the factorial of each repeated count so identical swaps are not counted twice:

\frac{6!}{4! \times 2!} = \frac{720}{24 \times 2} = 15

Answer

There are 15 different codes. Sense check: swapping the 4 P's among themselves ( $4!$

ways) or the 2 Q's among themselves ( $2!$ ways) never produces a new-looking code, so dividing $720$ by $4!\times 2!$ correctly removes that overcounting.

Q6[5 marks]

Seven sprinters are to be arranged in a row of 7 lanes for a practice run. Two of the sprinters, P and Q, must not be placed in adjacent lanes.

In how many ways can the 7 sprinters be arranged in the row?

Show worked solution

Rather than counting the allowed arrangements directly, count all arrangements and subtract the ones that break the rule. With 7 distinct sprinters in a row, order matters throughout, so first find the total number of arrangements:

7! = 5040

Now count the unwanted arrangements, where P and Q are adjacent. Glue P and Q into one block, giving 6 units to arrange in a row, and let the two sprinters swap places inside the block:

6! \times 2! = 720 \times 2 = 1440

Subtract the unwanted arrangements from the total, leaving only those where P and Q are not adjacent:

5040 - 1440 = 3600

Answer

There are 3600 valid arrangements. Sense check: in a row of 7 distinct people, two chosen people are adjacent in exactly $\frac{2}{7}$ of all orders, so the \u2018not adjacent\u2019 count should be $5040 \times \left(1 - \frac{2}{7}\right) = 5040 \times \frac{5}{7} = 3600$ \u2014 matching.

Q7[5 marks]

A team of 5 players is to be selected from 10 players to represent a school in a competition. Players A and B cannot both be selected into the same team, because their training schedules clash.

In how many ways can the team of 5 be selected?

Show worked solution

The condition \u2018A and B cannot both be in the team\u2019 splits the selection into three cases that cannot happen together: neither A nor B is chosen, only A is chosen, or only B is chosen. Count each case and add them.

Case 1 \u2014 neither A nor B: choose all 5 players from the remaining 8:

{}^{8}C_{5} = 56

Case 2 \u2014 A is chosen but B is not: A already fills one place, so choose the remaining 4 players from the other 8 players (excluding B):

{}^{8}C_{4} = 70

Case 3 \u2014 B is chosen but A is not, which by the same reasoning also gives ${}^{8}C_{4} = 70$ . The three cases are mutually exclusive, so add all three:

56 + 70 + 70 = 196

Answer

There are 196 valid teams. Check by complement: choosing any 5 from 10 gives ${}^{10}C_{5} = 252$ teams in total, and the unwanted teams containing both A and B need 3 more players from the remaining 8, giving ${}^{8}C_{3} = 56$ ; so $252 - 56 = 196$ \u2014 the same answer.

Q8[6 marks]

8 students in a class are to be divided into two groups of 4 to travel in two identical vans for a class outing \u2014 the vans are not numbered in any way. In how many ways can the 8 students be divided into the two groups?

Show worked solution

This is a division into groups, not a straightforward selection, because the two vans are identical \u2014 there is no \u2018first group\u2019 or \u2018second group\u2019 to tell them apart. Start as if choosing one group of 4 from the 8 students, which automatically leaves the remaining 4 for the other van:

{}^{8}C_{4} = 70

This counts every division twice. Choosing students $\{1,2,3,4\}$ as \u2018the chosen group\u2019 leaves $\{5,6,7,8\}$ for the other van, and choosing $\{5,6,7,8\}$ as \u2018the chosen group\u2019 leaves $\{1,2,3,4\}$ \u2014 but since the vans are identical, these describe the exact same division of students.

Every division has been counted exactly twice, so divide by $2!$ :

\frac{70}{2!} = \frac{70}{2} = 35

This halving step is only needed because the two groups play an identical role. If the vans were instead numbered Van 1 and Van 2, the two selections above would count as different outcomes, and dividing by $2!$

would not apply.

Answer

There are 35 ways to divide the students. Sense check: for a smaller case of 4 students split into two identical groups of 2, the same method gives $\frac{{}^{4}C_{2}}{2!} = \frac{6}{2} = 3$ , which matches the 3 pairings you can list by hand, confirming the halving step is genuine.

Across these three, the decisive skill is not the formula but the plan: fix the tightest condition first, add when the cases are alternatives, and multiply when the stages combine. Write that plan in words before the numbers, and even a long question stays under control.

Key method points

These three examples rehearse the planning skills that separate a secure hard-question answer from a lucky one. Keep the following points in mind as you practise more.

For number-forming problems, satisfy the most restricted position first (often the units digit for even or odd conditions), then fill the rest.
‘At least’ and ‘at most’ conditions split into separate cases; count each case and add the results.
Independent stages of one outcome are multiplied; mutually exclusive cases are added, mixing these up is the most common hard-question error.
A ‘select then arrange’ task needs both tools: a combination to choose the group, then a permutation to order it.
Always check whether order matters separately at each stage, it can matter in one stage and not another.
Because marking is analytic, each correctly counted case or stage earns method marks even before the final addition or multiplication.

How a teacher helps

Hard permutation questions are rarely lost on arithmetic; they are lost on the plan, adding when cases should be multiplied, or arranging when the question only asked to select. In a one-to-one lesson our teacher builds the plan first, writing the cases and stages in words before a single number appears, so the structure is right from the start.

Because our teachers are experienced, you learn a repeatable way to break a long question down, not a trick for one question. Lessons are taught in English, while SPM papers are set in both Malay and English, so the same plan works on either paper.

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Frequently asked questions

When do I add cases and when do I multiply?

Add when the cases are alternatives that cannot happen together, such as ‘exactly 3 women’ or ‘exactly 4 women’. Multiply when independent stages combine to build one outcome, such as choosing the men and choosing the women for the same team.

Why fix the units digit first when forming even numbers?

Because ‘even’ is a condition on the units digit only, and it is the most restricted position. Fixing it first, here $3$ even choices, leaves the freer positions to be filled afterwards without breaking the rule.

In ‘select then arrange’, why do I use both a combination and a permutation?

Choosing who is on the panel is a selection where order does not matter (combination). Seating the chosen members in a row is an arrangement where order does matter (permutation).

Because the two stages are independent, their counts are multiplied: $30 \times 6 = 180$ .

How can I check an ‘at least’ answer?

One way is the complement: count everything and subtract the unwanted cases. Another, used here, is to list the allowed cases directly and add them.

If both routes agree, the answer is secure.

Source:SRC-DSKP-EN

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

Permutation and Combination, Worked Examples (KBAT)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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