 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-3/6d/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The orbit of

(1, \dots, k)

consists precisely of the ordered

k

{1, \dots, n}

Orb (x) = {(x_{1}, \dots, x_{k}) : x_{i} \neq = x_{j} for i \neq = j} .

It has size

n (n - 1) \dots (n - k + 1) = \frac{n !}{( n - k )!} .

The stabilizer consists of permutations fixing

1, \dots, k

pointwise, while freely permuting the remaining letters:

Stab (x) ≅ S_{n - k}, ∣ Stab (x) ∣ = (n - k)! .

Thus

∣ Orb (x) ∣ ∣ Stab (x) ∣ = \frac{n !}{( n - k )!} (n - k)! = n! = ∣ S_{n} ∣,

verifying orbit-stabilizer.

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