Find the total number of different arrangements of the 11 letters in the word CATERPILLAR.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2021). The mark or mark type is shown in brackets.

1. [1] M1 11! alone as numerator. 2! × m!× n! on denominator, m = 1, 2, n = 1, 2. no additional terms, no additional operations.
2. [1] A1 Exact answer only.
3. [1] B1 7!/2! x k seen, k an integer > 1.
4. [1] M1 m×n(n-1) or m×ⁿC₂ or m×ⁿP₂, n = 7, 8 or 9, m an integer > 1.
5. [1] M1 7!/2! x ⁸C₂, or 7!/p! x ⁸P₂, p integer ≥ 1, condone 2520×28.
6. [1] A1 Exact answer only. SC B1 70560 from M0, M1 only.
7. [1] M1 9! / (2!m!) k, 90720 > k integer > 1, m = 1, 2.
8. [1] B1 8! / s, s an integer >1
9. [1] M1 9!/p - 8!/q , p, q integers ≥ 1, condone 90720 – 20160.
10. [1] A1 Exact answer only. SC B1 70560 from M0, M1 only.
11. [1] M1 ⁿCₓ seen alone or ⁿCₓ × k, 2≥ k ≥ 1, k an integer, 0 < x < 5 linked to an appropriate scenario.
12. [1] A1 ⁵C₂ x k, k = 1 oe or ⁵C₁ × m, m = 1,2 oe alone. SC if ⁿCₓ not seen. B2 for 5 or 10 linked to the appropriate scenario WWW.
13. [1] M1 Add outcomes from 3 or 4 identified correct scenarios only, accept unsimplified. 2Cw x 2Cx x 2Cy × 5Cz, w+x+y+z=6 identifies w Rs, × As and y Ls.
14. [1] A1 WWW, only dependent on 2nd M mark. Note: ⁵C₂ + ⁵C₁ + ⁵C₁ + 1 = 21 is sufficient for 4/4.
15. [1] SC not all (or no) scenarios identified. B1 10+5+5+1 DB1 = 21
16. [1] M1 ⁿCₓ seen alone or ⁷Cₓ × k, 2≥ k≥ 1, k an integer, 0<x<7. Condone ⁿPₓ or ⁿPₓ × k, 2≥ k≥ 1, k an integer, 0<x<7.
17. [1] M1 ⁷C₂ × k, 2≥ k≥ 1oe
18. [1] A1 ⁷C₂× k, k = 1oe no other terms.
19. [1] A1 Value stated.