Solve arrangement and selection problems in context using permutations or combinations

Permutations and Combinations – Cambridge IGCSE Additional Mathematics (0606)

Objective

To solve arrangement and selection problems in context by deciding whether a permutation or a combination is required and applying the appropriate formula.

Syllabus Alignment (Section 11)

• 11.1 Recognise the difference between permutations and combinations – Identify when order matters (permutation) and when it does not (combination).
• 11.2 Know and use the notation \(n!\) and the expressions for permutations/combinations of n items taken r at a time – Use both the Cambridge notation \(nP_r,\;nC_r\) and the alternative symbols \({}^{n}P_{r},\;{^{n}C_{r}}\).
• 11.3 Solve arrangement and selection problems in context – Apply the formulas, interpret the answer and check that it is reasonable.

Quick‑Check: Permutation vs Combination

Key Concepts

• Permutation: An ordered arrangement of objects; the sequence matters.
• Combination: A selection of objects where the order does **not** matter.
• Factorial:

  \(n! = n \times (n-1) \times \dots \times 2 \times 1\) for \(n\ge 1\)

  \(0! = 1\) (by definition)

  Useful identity: \((n+1)! = (n+1)\,n!\)

• No repetitions: The IGCSE 0606 syllabus for this section does **not** cover arrangements where items may be repeated. All formulas assume distinct objects.

Standard Notation

When to Use Permutations

Use a permutation when the problem asks for the number of possible ways to arrange or order a set of items. Typical keywords include:

• arrange, order, line‑up, seat, rank, first/second/third, position, sequence, password (when order matters)

When to Use Combinations

Use a combination when the problem asks for the number of ways to select a subset of items where the order is irrelevant. Look for words such as:

• choose, select, pick, form a committee, draw, hand, group, team, subset

Worked Examples

Example 1 – Permutation (Line‑up)

Problem: A school choir has 8 members. In how many different ways can the choir director choose a line‑up of 5 singers for a performance?

1. Order matters → permutation.
2. Use \({}^{8}P_{5}= \dfrac{8!}{(8-5)!}\).
3. \({}^{8}P_{5}= \dfrac{8!}{3!}= \dfrac{40320}{6}=6720.\)
4. Answer: 6 720** possible line‑ups.

Example 2 – Combination (Project Group)

Problem: From a class of 12 students, a teacher wants to form a group of 4 for a project. How many different groups can be formed?

1. Order does not matter → combination.
2. Use \({}^{12}C_{4}= \dfrac{12!}{4!\,8!}\).
3. \({}^{12}C_{4}= \dfrac{12\times11\times10\times9}{4\times3\times2\times1}=495.\)
4. Answer: **495** possible groups.

Example 3 – Mixed Situation (Committee + Round Table)

Problem: A club must choose a committee of 3 members from 7 people and then seat those 3 members round a circular table. How many different outcomes are possible?

1. Choosing the committee: combination \({}^{7}C_{3}=35.\)
2. Arranging 3 people round a circle: \((3-1)! = 2! = 2\) (circular permutation formula).
3. Total outcomes = \(35 \times 2 = 70.\)
4. Answer: **70** possible committee‑and‑seating arrangements.

Example 4 – Password (Two Separate Permutations)

Problem: A password consists of 4 distinct letters followed by 2 distinct digits. How many such passwords can be formed? (Assume 26 letters and 10 digits.)

1. Letters are ordered → \({}^{26}P_{4}=26\times25\times24\times23.\)
2. Digits are ordered → \({}^{10}P_{2}=10\times9.\)
3. Total passwords = \((26\times25\times24\times23)\times(10\times9)=358\,800.\)
4. Answer: **358 800** possible passwords.

Common Pitfalls

• Keyword mis‑identification – “arrange” → permutation; “choose” → combination.
• Missing denominator in permutations – Forgetting the \((n-r)!\) term leads to an over‑count.
• Over‑counting in combinations – Omitting the \(r!\) factor leaves order counted multiple times.
• Repeating items – This section does not cover repetitions; do not use formulas that allow repeated selections.
• Not checking reasonableness – Compare the answer with the total number of possible outcomes (e.g., \(n!\) for full arrangements) to spot obvious errors.

Practice Questions

1. A password consists of 4 distinct letters followed by 2 distinct digits. How many such passwords can be formed? (Assume 26 letters and 10 digits.)
2. From a standard deck of 52 playing cards, how many 5‑card hands contain exactly 3 hearts?
3. In how many ways can 7 books be placed on a shelf if 2 particular books must not be next to each other?
4. A committee of 3 boys and 2 girls is to be chosen from 8 boys and 7 girls. How many different committees are possible?
5. How many different ways can the letters of the word “LEVEL” be arranged?
6. From 9 applicants, a team of 4 is to be selected and then appointed as captain, vice‑captain, secretary and treasurer. How many possible outcomes are there?

Answer Key (for teacher use)

1. \({}^{26}P_{4}\times{}^{10}P_{2}= (26\times25\times24\times23)\times(10\times9)=358\,800.\)
2. Choose 3 hearts from 13: \({}^{13}C_{3}=286\).
   Choose the remaining 2 cards from the 39 non‑hearts: \({}^{39}C_{2}=741\).
   Total \(=286\times741=211\,746.\)
3. Total arrangements of 7 books: \(7! = 5040\).
   Arrangements with the two particular books together: treat them as a block → \(6!\times2! = 720\times2 = 1440\).
   Required arrangements \(=5040-1440=3600.\)
4. Choose boys: \({}^{8}C_{3}=56\). Choose girls: \({}^{7}C_{2}=21\).
   Total committees \(=56\times21=1176.\)
5. Word “LEVEL”: L appears 2, E appears 2, V appears 1.
   Distinct arrangements \(=\dfrac{5!}{2!\,2!}= \dfrac{120}{4}=30.\)
6. Select 4 people: \({}^{9}C_{4}=126\).
   Assign 4 distinct offices (order matters): \(4! = 24\).
   Total outcomes \(=126\times24=3024.\)

Suggested Diagram