19N.2.AHL.TZ0.H_8

pestleMathematicsAAHLPaper 219N· sl-1-9-binomial-theorem-where-n-is-an-integersource ↗

Eight boys and two girls sit on a bench. Determine the number of possible arrangements, given that

the girls do not sit together.

[3]

the girls do not sit on either end.

[2]

the girls do not sit on either end and do not sit together.

[3]

Markscheme / solution

METHOD 1

$10{\text{!}} - 2 \times 9{\text{!}}\,\left( { = 2903040} \right)$ (A1)(A1)A1

Note: Award A1 for $10{\text{!}}$ , A1 for $2 \times 9{\text{!}}$ , A1 for final answer.

METHOD 2

$9 \times 8 \times 8{\text{!}}$ (A1)(A1)A1

Note: Award A1 for $9 \times 8$ or equivalent, A1 for $8{\text{!}}$ and A1 for answer.

[3 marks]

METHOD 1

$8 \times 7 \times 8{\text{!}}\,\left( { = 2257920} \right)$ (A1)A1

Note: Award (A1) for $8 \times 7$ , A1 for final answer.

METHOD 2

$10{\text{!}} - 2 \times 8{\text{!}} - 2 \times 2 \times 7 \times 8{\text{!}}$

Note: Award A1 for $10{\text{!}}$ minus EITHER subtracted terms and A1 for final correct answer.

[2 marks]

METHOD 1

$8 \times 7 \times \left( {8{\text{!}} - 2 \times 7{\text{!}}} \right)\,\left( { = 1693440} \right)$ (A1)(A1)A1

Note: Award (A1) for $8 \times 7$ , (A1) for ${2 \times 7{\text{!}}}$ , A1 for final answer. $\left( {8{\text{!}} - 2 \times 7{\text{!}}} \right)$ can be replaced by $6 \times 7{\text{!}}$ or ${}^7{P_2} \times 6{\text{!}}$ which may be awarded the second A1.

METHOD 2

their answer to (a) $- 2 \times 8{\text{!}} - 2 \times 2 \times 7 \times 8{\text{!}}$ (A1)(A1)A1

Note: Award A1 for subtracting each of the terms and A1 for final answer.

METHOD 3

their answer to (b) $- 2 \times 7 \times 8{\text{!}}$ or equivalent (A1)A2

Note: Award A1 for the subtraction and A2 for final answer.

[3 marks]

Examiners’ report

[N/A]