A car park has spaces for 18 cars, arranged in a line. On one day there are 5 cars, of different makes, parked in randomly chosen positions and 13 empty spaces.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

1. [M1] 18P5 or 18Px or xP5 OE seen, 0 < x < 18 and 5 <y <18, can be mult by k ≥ 1
2. [A1] 1 028 160
3. [B1] 5! OE mult by k ≥ 1, considering the arrangements of cars next to each other
4. [B1] Mult by 14 OE, (or 14 on its own) considering positions within the line
5. [M1] Dividing by (i) for probability
6. [M1] Subtracting prob from 1 (or their '5! × 14' from (i))
7. [A1] = 0.998 (611/612) OE
8. [B1] 5! OE mult by k ≥ 1 (on its own or in numerator of fraction) considering the arrangements of cars next to each other
9. [B1] Multiply by 14!, (or 14! on its own) considering all ways of arranging spaces with 5 cars together
10. [M1] Dividing by 18!, total number of ways of arranging spaces
11. [M1] Subtracting prob from 1 (or '5! × 14!' from 18!)
12. [A1] = 0.998(366)
13. [M1] Listing the six correct scenarios (only): 4 together; 3 together and 2 separate; 3 together and 2 together; two sets of 2 together and 1 separate; 2 together and 3 separate; 5 separate.
14. [M1] Summing total of the six scenarios, at least 2 correct unsimplified
15. [A1] Total = 1 026 480
16. [M1] Dividing their 1 026 480 by their 6(i)
17. [A1] 1026 480 / 1028160 = 0.998(366)
18. [B1] 5C1×4C1×3C1 or better seen i.e. no. of ways with 3 different colours
19. [M1] Any of 5C2 or 4C2 or 3C2 seen multiplied by k≥ 1 (can be implied)
20. [A1] 2 correct unsimplified 'no. of ways' other than 5C1 × 4C1 × 3C1
21. [M1] Summing no more than 7 scenario totals containing at least 6 correct scenarios
22. [A1] Total = 205
23. [M1] Seeing 12C3 –', considering all selections of 3 cars
24. [M1] Subt 5C3 OE, removing only red selections
25. [M1] Subt 4C3 OE, removing only white selections
26. [M1] Subt 3C3 OE, removing only black selections
27. [A1] Correct answer