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A-LevelMathematicsProbability and statistics9709/53 · Oct/Nov 2023 · Q610 Marks

Jai and his wife Kaz are having a party. Jai has invited five friends and each friend will bring his wife.

  1. (a)(i)At the beginning of the party, the 12 people will stand in a line for a photograph. How many different arrangements are there of the 12 people if Jai stands next to Kaz and each friend stands next to his own wife?[3]
  2. (a)(ii)How many different arrangements are there of the 12 people if Jai and Kaz occupy the two middle positions in the line, with Jai's five friends on one side and the five wives of the friends on the other side?[2]
  3. (b)For a competition during the party, the 12 people are divided at random into a group of 5, a group of 4 and a group of 3. Find the probability that Jai and Kaz are in the same group as each other.[5]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

  1. [M1] 6! × a, a integer > 1.
  2. [M1] b × 2⁶, b integer ≥ 1.
  3. [A1] Accurate answer required. SC B1 for 46080 if M0 M0 www.
  4. [M1] c × d × e × f × g × h 2 < c, d, e, f, g, h (different integers) < 12.
  5. [M1] Correct unsimplified.
  6. [A1] Accurate answer required. SC B1 for 46080 if M0 M0 www.
  7. [M1] 5! × 5! × k, k positive integer, 1 may be implied (no adding/subtracting).
  8. [A1]
  9. [B1] Correct probability for one identified scenario. In the group of 5: 5/12 × 4/11 = 20/132 [= 5/33] In the group of 4: 4/12 × 3/11 = 12/132 [= 1/11] In the group of 3: 3/12 × 2/11 = 6/132 [= 1/22]
  10. [M1] Denominator 12×11 for all probabilities, (1, 2 or 3 scenarios).
  11. [A1] 3 correct probabilities, accept unsimplified.
  12. [M1] Adding probabilities for 3 correct scenarios.
  13. [A1] 0.2878787 to at least 3SF.
  14. [B1] Correct value of one identified scenario seen, accept unsimplified. In the group of 5: ¹⁰C₃ × ⁷C₄ [= 120 × 35 = 4200] In the group of 4: ¹⁰C₂ × ⁸C₅ [= 45 × 56 = 2520] In the group of 3: ¹⁰C₁ × ⁹C₅ [= 10 × 126 = 1260]
  15. [M1] ¹²Cₐ × ¹²⁻ᵃCь, a = 3, 4, 5; b = 3, 4, 5 (a ≠ b)
  16. [A1] 27720 Seen alone or as denominator of probability - accept unsimplified. SC B1 if M0.
  17. [M1] Values of 3 correct scenarios added, accept unsimplified – or correct.
  18. [A1] 0.2878787 to at least 3SF.

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About this question

Topic
Probability and statistics, A-Level Mathematics (syllabus 9709). Structured question, 10 marks.
Source
A-Level Mathematics, Oct/Nov 2023 Paper 5 Variant 3, question 6.

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