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Date	May 2022	Marks available	3	Reference code	22M.2.AHL.TZ1.9
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Find	Question number	9	Adapted from	N/A

Question

Mary, three female friends, and her brother, Peter, attend the theatre. In the theatre there is a row of $10$ empty seats. For the first half of the show, they decide to sit next to each other in this row.

For the second half of the show, they return to the same row of $10$ empty seats. The four girls decide to sit at least one seat apart from Peter. The four girls do not have to sit next to each other.

Find the number of ways these five people can be seated in this row.

[3]

Find the number of ways these five people can now be seated in this row.

[4]

Markscheme

$6 \times 5!$ (A1)(A1)

$= 720$ (accept $6!$ ) A1

[3 marks]

METHOD 1

(Peter apart from girls, in an end seat) ${}^{8}P_{4} (= 1680)$ OR

(Peter apart from girls, not in end seat) ${}^{7}P_{4} (= 840)$ (A1)

case 1: Peter at either end

$2 \times {}^{8}P_{4} (= 3360)$ OR $2 \times {}^{8}C_{4} \times 4! (= 3360)$ (A1)

case 2: Peter not at the end

$8 \times {}^{7}P_{4} (= 6720)$ OR $8 \times {}^{7}C_{4} \times 4! (= 6720)$ (A1)

Total number of ways $= 3360 + 6720$

$= 10080$ A1

METHOD 2

(Peter next to girl, in an end seat) $4 \times {}^{8}P_{3} (= 1344)$ OR

(Peter next to one girl, not in end seat) $2 \times 4 \times {}^{7}P_{3} (= 1680)$ OR

(Peter next to two girls, not in end seat) $4 \times 3 \times {}^{7}P_{2} (= 504)$ (A1)

case 1: Peter at either end

$2 \times 4 \times {}^{8}P_{3} (= 2688)$ (A1)

case 2: Peter not at the end

$8 (2 \times 4 \times {}^{7}P_{3} + 4 \times 3 \times {}^{7}P_{2}) (= 17472)$ (A1)

Total number of ways $= {}^{10}P_{5} - (2688 + 17472)$

$= 10080$ A1

[4 marks]

Examiners report

[N/A]

Syllabus sections

Topic 1—Number and algebra » AHL 1.10—Perms and combs, binomial with negative and fractional indices

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