Practice Maths

Permutations

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Key Terms

Permutation P(n, r)
The number of ordered arrangements of r objects chosen from n distinct objects; P(n,r) = n!/(n−r)!.
All n objects
P(n, n) = n! arrangements of n distinct objects.
Repeated elements
n! / (n1! × n2! × …) where ni is the count of each repeated element.
Order matters
Permutations count ordered sequences; ABC and BAC are different permutations but the same combination.
Restricted arrangements
Treat fixed elements as a unit, or use case analysis to enumerate valid arrangements.
Circular permutations
(n−1)! arrangements of n objects in a circle (one position is fixed to remove rotational symmetry).

Permutation Formulas

Ordered arrangements of r objects from n distinct objects:

P(n, r) = n! / (n − r)! = n × (n−1) × … × (n−r+1)

Arrangements of all n distinct objects:   n!

Arrangements with repeated elements:

n! / (n1! × n2! × …)    where ni is the count of each repeated element

SituationUseKey Question
Arranging objects in a sequencePermutation — order mattersDoes position/rank matter?
Choosing a group with no specific rolesCombination — order doesn’t matterIs ABC the same as BAC?

Worked Example 1 — P(8, 3)

How many ways can 3 people from a group of 8 be arranged in order (1st, 2nd, 3rd)?

P(8, 3) = 8 × 7 × 6 = 336

Equivalently: 8! / (8−3)! = 8! / 5! = 336

Worked Example 2 — Repeated Letters

How many distinct arrangements of the letters in ARRANGE are there?

Letters: A(2), R(2), N(1), G(1), E(1) → total 7 letters with A repeated twice and R repeated twice.

Arrangements = 7! / (2! × 2!) = 5040 / 4 = 1 260

Hot Tip: Permutations count ordered arrangements — position matters. If the question uses words like “rank”, “arrange”, “order”, or “sequence”, it is a permutation. If it says “choose a group” or “select a team” where order does not matter, it is a combination. When letters repeat, divide by the factorial of each repeated count.

Why P(n, r) = n! / (n−r)!

To fill r positions from n distinct objects: position 1 has n choices, position 2 has n−1, …, position r has n−r+1 choices. By the multiplication principle:

P(n, r) = n × (n−1) × … × (n−r+1)

Multiplying numerator and denominator by (n−r)! converts this to n!/(n−r)!, which is the standard formula.

Arrangements with Restrictions

When some elements must occupy specific positions, fix the restricted elements first, then count freely for the remaining positions. This avoids accidentally counting invalid arrangements.

Example: 6 people in a row, A must be at one end. Fix A in position 1 or position 6 (2 ways), then arrange the other 5 in the remaining 5 positions (5! = 120 ways). Total = 2 × 120 = 240.

Adjacent Elements

When two elements must be adjacent (next to each other), treat them as a single unit. If n objects are arranged with 2 specific objects always together, think of it as (n−1) units to arrange: (n−1)! arrangements of units × 2! internal orderings of the pair.

Arrangements with Repeated Elements

If n objects contain n1 identical items of type 1, n2 of type 2, etc., the number of distinct arrangements is n! divided by the product of the factorials of each repeated count. Division removes the overcounting caused by swapping identical items.

Why divide? Swapping two identical letters (e.g., the two S's in MISSISSIPPI) creates no new arrangement, but the raw n! count treats each swap as different. Dividing by the factorial of each repeat count removes this overcounting.

Circular Arrangements (Enrichment)

When n objects are arranged in a circle, one position can be fixed to remove rotational equivalence. This gives (n−1)! distinct circular arrangements instead of n!.

Mastery Practice

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  1. Fluency

    Q1 — Calculate Permutations

    Calculate:   (a) P(6, 2)    (b) P(9, 3)    (c) P(5, 5)

  2. Fluency

    Q2 — Seating Arrangements

    In how many ways can 8 people be seated in a row of 8 chairs?

  3. Fluency

    Q3 — 3-Letter Arrangements

    How many different 3-letter arrangements can be made from the 7 distinct letters {A, B, C, D, E, F, G}, using each letter at most once?

  4. Fluency

    Q4 — Selecting Office-Bearers

    In how many ways can a president, vice-president and secretary be selected from a group of 12 people?

  5. Understanding

    Q5 — Arranging Letters

    How many different arrangements are there of the letters in:   (a) MATHS    (b) MISSISSIPPI?

  6. Understanding

    Q6 — Row Seating with Restriction

    Six people sit in a row. In how many ways can this be done if:   (a) there are no restrictions?    (b) Ann and Bob must sit next to each other?

  7. Understanding

    Q7 — 5-Digit Numbers from {1, 2, 3, 4, 5}

    A 5-digit number is formed using the digits 1, 2, 3, 4, 5, each exactly once. How many such numbers:   (a) are there in total?    (b) are even?    (c) have the digit 3 in the middle position?

  8. Understanding

    Q8 — Arrangements of PARALLEL

    How many distinct arrangements of the letters in the word PARALLEL are there?

  9. Problem Solving

    Q9 — Bookshelf Arrangements

    A bookshelf holds 4 different Maths books, 3 different Science books and 2 different English books. In how many ways can they be arranged if:   (a) there are no restrictions?    (b) all books of the same subject must be kept together?

  10. Problem Solving

    Q10 — 4-Digit Numbers with Leading Digit Restriction

    The digits 0, 1, 2, 3, 4, 5 are used without repetition to form 4-digit numbers. How many such 4-digit numbers are there? (Note: 0 cannot be the leading digit.)

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