CIE AS/A Level Maths-5.2 Permutations and combinations- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-5.2 Permutations and combinations- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-5.2 Permutations and combinations- Study Notes
Key Concepts:
- Permutations and Combinations
- Arrangements in a Line – Repetition and Restrictions
Permutations and Combinations
Permutations and Combinations
• A permutation refers to the arrangement of objects in a specific order. The order of selection matters.
Formula: \( ^nP_r = \dfrac{n!}{(n-r)!} \)
• A combination refers to the selection of objects without considering the order. The order of selection does not matter.
Formula: \( ^nC_r = \dfrac{n!}{r!(n-r)!} \)
• Relationship: \( ^nP_r = ^nC_r \times r! \)
• Key differences:
- Permutations: Order matters (e.g., arranging books on a shelf).
- Combinations: Order does not matter (e.g., choosing members of a team).
• Factorial Definition: \( n! = n \times (n-1) \times (n-2) \times \dots \times 1 \). Special case: \( 0! = 1 \).
Example:
How many ways can 4 students be arranged in a row of 4 chairs?
▶️ Answer/Explanation
Total students \( n = 4 \), number of chairs \( r = 4 \).
Number of arrangements = \( ^4P_4 = \dfrac{4!}{(4-4)!} = 4! = 24 \).
Final Answer: \(\boxed{24}\) ways.
Example :
A committee of 3 members is to be chosen from 5 people. In how many ways can this be done?
▶️ Answer/Explanation
Here \( n = 5, r = 3 \).
Number of selections = \( ^5C_3 = \dfrac{5!}{3!(5-3)!} = \dfrac{5 \times 4 \times 3!}{3! \times 2!} = 10 \).
Final Answer: \(\boxed{10}\) ways.
Arrangements in a Line — Repetition and Restrictions
Arrangements in a Line — Repetition and Restrictions
When arranging \(n\) distinct objects in a line, the number of arrangements is
\(n!\).
Repetition / identical items:
If some objects are identical (repeated), count distinct arrangements by dividing by the factorials of identical counts:
\(\text{Number of distinct arrangements}=\dfrac{n!}{n_1!\,n_2!\cdots}\)
where \(n\) is total objects and \(n_1,n_2,\dots\) are multiplicities of identical objects.
Restriction — two particular people must stand together:
Treat the two people who must be adjacent as a single compound object, then multiply by the internal arrangements of that pair:
If pair treated as one, arrangements = \((n-1)!\times 2!\)
Restriction — two particular people must not stand together:
Count total arrangements and subtract arrangements where they are together:
\(\text{Not together} = n! – (\text{together count})\)
More complex restrictions:
Use inclusion–exclusion, treat blocks, or case-work (split into mutually exclusive cases) as appropriate.
Example
How many distinct linear arrangements (words) can be made from the letters of the word NEEDLESS?
▶️ Answer/Explanation
Step 1: Count letters
The word NEEDLESS has 8 letters in total. Letter multiplicities: N:1, E:3, D:1, L:1, S:2.
Step 2: Use permutation-with-repetition formula
Number \(= \dfrac{8!}{1!\,3!\,1!\,1!\,2!}.\)
Compute factorials: \(8! = 40320,\ 3! = 6,\ 2! = 2.\)
Number \(= \dfrac{40320}{6\times 2} = \dfrac{40320}{12} = 3360.\)
\(\boxed{3360}\)
Example
Five people A, B, C, D, E stand in a line. In how many ways can they line up if A and B must stand next to each other?
▶️ Answer/Explanation
Step 1: Treat A and B as a single block
Then we are arranging the block (AB) plus C, D, E → total \(4\) objects.
Step 2: Arrangements of the 4 objects
There are \(4!\) ways to order them.
Step 3: Internal arrangements of A and B within the block
A and B can be arranged in \(2!\) ways (AB or BA).
Step 4: Multiply
Total \(=4!\times2! = 24\times2 = 48.\)
\(\boxed{48}\)
Example
Using the same five people A, B, C, D, E, how many arrangements are there if A and B must not stand next to each other?
▶️ Answer/Explanation
Step 1: Compute total arrangements with no restriction
Total \(=5! = 120.\)
Step 2: Compute arrangements where A and B are together
From Example 2, together count \(=48.\)
Step 3: Subtract
Not together \(= 120 – 48 = 72.\)
\(\boxed{72}\)
Example
Seven people: A, B, C, D, E, F, G. Pairs (A,B) and (C,D) must each stand together (each pair adjacent). Find the number of arrangements.
▶️ Answer/Explanation
Step 1: Treat each required adjacent pair as a block
Blocks: (AB), (CD), plus E, F, G → total objects \(=5\).
Step 2: Arrange the 5 objects
Ways \(=5!.\)
Step 3: Internal arrangements within each block
Each pair can be ordered in \(2!\) ways, so multiply by \(2!\) for (AB) and \(2!\) for (CD).
Step 4: Multiply together
Total \(=5!\times2!\times2! = 120\times2\times2 = 480.\)
\(\boxed{480}\)
Useful tips
- Always decide whether order matters (permutations) or not (combinations) — for linear arrangements order matters.
- When identical items appear, divide by factorial(s) for repeated counts.
- For adjacency restrictions, consider block method (treat adjacent group as a single object) and remember to multiply by internal permutations of the block.
- For “must not be together”, use total minus together. For multiple restrictions, consider inclusion–exclusion or careful case splitting.
- Check whether additional symmetries or identical people reduce counts further (e.g. identical twins → divide by 2!).
Example
Six students are to be seated in two rows of three chairs each. In how many different ways can they be arranged?
▶️ Answer/Explanation
Step 1: There are 6 students and 6 seats (arranged in 2 rows of 3).
Step 2: The total number of permutations = \(6! = 720\).
Step 3: Since the rows are distinguishable (front row vs back row), no adjustment is required.
Final Answer: \(\boxed{720}\) ways.
Example
Eight people are to be seated in two rows of four chairs each. In how many ways can they be arranged if two particular friends must sit in the same row?
▶️ Answer/Explanation
Step 1: Choose the row for the two friends: \(2\) ways (front or back row).
Step 2: Choose 2 out of 4 seats in that row: \(\binom{4}{2} = 6\) ways.
Step 3: Arrange the two friends in those 2 seats: \(2!\) ways.
Step 4: Arrange the remaining 6 people in the 6 remaining seats: \(6!\) ways.
Total arrangements = \(2 \times 6 \times 2! \times 6! = 2 \times 6 \times 2 \times 720 = 17{,}280\).
Final Answer: \(\boxed{17{,}280}\) ways.