A-LevelMathematicsProbability and statistics9709/61 · May/June 2019 · Q89 Marks

Freddie has 6 toy cars and 3 toy buses, all different. He chooses 4 toys to take on holiday with him.

(i)In how many different ways can Freddie choose 4 toys?[1]
(ii)How many of these choices will include both his favourite car and his favourite bus?[2]
(iii)Freddie arranges these 9 toys in a line. Find the number of possible arrangements if the buses are all next to each other.[3]
(iv)Find the number of possible arrangements if there is a car at each end of the line and no buses are next to each other.[3]

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

[1] (9C4 =) 126. B1
[2] (6C1 × 3C1 × 5C2) / 6C1 × 3C1 × 5C2 or 6C2 or 3C2 (implied by correct answer) or 6P1 or 3P1 or 5P2, seen alone B1 = 21. correct answer B1
[3] (9! / (6! × 3!) × 3! × 6!) or (7! × 3!) or (6! × 3! × 7) = 30240. 3! or 6! seen alone or multiplied by k > 1 need not be an integer B1 3! × 6! × 7. 3! and 6! seen multiplied by k > 1, integer, no division B1 = 30240. Exact value B1 Alternative method for question 8(iii) C1 (B1 B2 B3) C2 C3 C4 C5 C6. 3! or 7! seen alone or multiplied by k > 1 need not be an integer B1 3! × 7!. 3! and 7! seen multiplied by k > or = 1, no division B1 = 30240. Exact value B1
[3] C1_C2_C3_C4 _ C5 _C6. 6! or 4! X 6P2 seen alone or multiplied by k > 1, no division (arrangements of cars) B1 6! × 5P3 or 6! × 5 × 4 × 3 or 6! x 3! x10. Multiply by 5P3 oe i.e. putting Bs in between 4 of the Cs OR multiply by 3! x n where n = 7, 8, 9, 10 (number of options) B1 = 43200. Correct answer B1

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