A-Level Mathematics October/November 2024 Q5(b): On another occasion a group of 5 will be chosen from the 21 students. The group must co…

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  Let V, G, and P be the number of violinists, guitarists, and pianists respectively. From the context of the question, we have V=10, G=6, P=5. Total students = 21.

  We need to choose a group of 5 students with the following conditions:

  • At least 2 violinists (V ≥ 2)
  • At least 1 guitarist (G ≥ 1)
  • At most 1 pianist (P ≤ 1)

  We can break this down into mutually exclusive cases based on the number of pianists (P=0 or P=1).

  Case 1: 1 Pianist is chosen (P=1) We need to choose 4 other students from violinists and guitarists. Let the number of violinists be $v$ and guitarists be $g$. We have $v+g=4$, with $v \ge 2$ and $g \ge 1$.

  • Subcase (i): 3 Violinists, 1 Guitarist, 1 Pianist Number of ways = $^{10}C_3 \times ^6C_1 \times ^5C_1 = 120 \times 6 \times 5 = 3600$
  • Subcase (ii): 2 Violinists, 2 Guitarists, 1 Pianist Number of ways = $^{10}C_2 \times ^6C_2 \times ^5C_1 = 45 \times 15 \times 5 = 3375$

  Case 2: 0 Pianists are chosen (P=0) We need to choose 5 students from violinists and guitarists. We have $v+g=5$, with $v \ge 2$ and $g \ge 1$.

  • Subcase (iii): 4 Violinists, 1 Guitarist, 0 Pianists Number of ways = $^{10}C_4 \times ^6C_1 \times ^5C_0 = 210 \times 6 \times 1 = 1260$
  • Subcase (iv): 3 Violinists, 2 Guitarists, 0 Pianists Number of ways = $^{10}C_3 \times ^6C_2 \times ^5C_0 = 120 \times 15 \times 1 = 1800$
  • Subcase (v): 2 Violinists, 3 Guitarists, 0 Pianists Number of ways = $^{10}C_2 \times ^6C_3 \times ^5C_0 = 45 \times 20 \times 1 = 900$

  Total Number of Ways The total number of ways is the sum of the ways for all possible cases. Total ways = $3600 + 3375 + 1260 + 1800 + 900$ Total ways = $10935$

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  How the marks are awarded

  • M1 — Identifying one correct scenario and writing a correct product of combinations, for example, $^{10}C_3 \times ^6C_1 \times ^5C_1$. The upper numbers (10, 6, 5) must be correct and the lower numbers (3, 1, 1) must sum to 5.
  • B1 — Correctly evaluating at least two of the identified scenarios. For example, calculating $3600$ for the (3V, 1G, 1P) case and $3375$ for the (2V, 2G, 1P) case. Unsimplified forms like $120 \times 6 \times 5$ are also accepted.
  • M1 — Adding the values for all 5 correct and distinct scenarios. The line 'Total ways = 3600 + 3375 + 1260 + 1800 + 900' earns this mark. No incorrect or repeated scenarios should be included.
  • A1 — Obtaining the correct final answer of 10935.

  Common mistakes

  • Misinterpreting 'at most 1 pianist' as 'exactly 1 pianist', which would lead to ignoring the three scenarios where no pianist is chosen.
  • Forgetting one of the five possible valid combinations of students, leading to an incomplete sum and a lower final answer.
  • Using permutations ($^nP_r$) instead of combinations ($^nC_r$). The order of selection for the group does not matter, so combinations are required.
  • Making a calculation error in one of the combinations, for example, miscalculating $^{10}C_4$ or $^{6}C_3$, which affects one of the subtotals and the final answer.

  Examiner tip: For complex selection problems with multiple constraints, systematically break the problem down into smaller, mutually exclusive cases, calculate each one, and then sum the results.

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