A-LevelMathematicsProbability and statistics9709/62 · May/June 2018 · Q610 Marks

(a) Find the number of ways in which all 9 letters of the word AUSTRALIA can be arranged in each of the following cases. (i) All the vowels (A, I, U are vowels) are together. (ii) The letter T is in the central position and each end position is occupied by one of the other consonants (R, S, L). (b) Donna has 2 necklaces, 8 rings and 4 bracelets, all different. She chooses 4 pieces of jewellery. How many possible selections can she make if she chooses at least 1 necklace and at least 1 bracelet?

(a)(i)All the vowels (A, I, U are vowels) are together.[3]
(a)(ii)The letter T is in the central position and each end position is occupied by one of the other consonants (R, S, L).[3]
(b)Donna has 2 necklaces, 8 rings and 4 bracelets, all different. She chooses 4 pieces of jewellery. How many possible selections can she make if she chooses at least 1 necklace and at least 1 bracelet?[4]

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

[M1] (a)(i) k x 5! (k is an integer, k ≥ 1)
[M1] (a)(i) m! / 3! (m is an integer, m ≥ 1)
[A1] (a)(i) Correct answer = 2400
[M1] (a)(ii) k x 6! / 3! or k x ³P₂ or k x ³C₂ or k × 3! or k × 3 × 2 (k is an integer, k ≥ 1), no irrelevant addition
[M1] (a)(ii) Correct unsimplified expression or 6!/3! x 3C2
[A1] (a)(ii) Correct answer = 720 ways
[M1] (b) Multiply 3 combinations, ²Cₓ×⁸Cᵧ×⁴Cᵥ. Accept ²C₁ = 2 etc.
[A1] (b) 3 or more options correct unsimplified
[M1] (b) Summing their values of 4 or 5 legitimate scenarios (no extra scenarios)
[A1] (b) Correct answer = 366

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