Council Committee, Photo, and Two Offices
A student council has 12 members: 7 in grade 12, including Ava and Ben, and 5 in grade 11, including Cara and Dev. The principal asks for a 4-member fundraising committee that includes at least one member from each grade. At the meeting, a council member proposes the count 7 x 5 x C(10,2) = 1575. Later, the five grade-11 members must be photographed in a single row for the yearbook, but Cara and Dev have asked not to stand side by side. Finally, whatever committee is formed must choose a chair and a secretary from among its own four members, and the principal wants the total number of possible outcomes reported alongside the committee count.
1. Is the proposed count of 1575 correct? Determine the correct number of 4-member committees with at least one member from each grade, and confirm it a second way.
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The count 1575 is far too large because it double counts. The factors 7 and 5 designate one grade-12 member and one grade-11 member as if they were distinguished positions, then C(10,2) picks two more from everyone left. A committee such as {Ava, Ben, Cara, Dev} is produced twice: once with Ava and Cara designated and Ben and Dev added, and once with Ben and Dev designated and Ava and Cara added.
Correct complement method: total committees minus the all-one-grade ones. C(12,4) - C(7,4) - C(5,4) = 495 - 35 - 5 = 455
Check by cases on the split: 1 and 3: C(7,1)C(5,3) = 7 × 10 = 70 2 and 2: C(7,2)C(5,2) = 21 × 10 = 210 3 and 1: C(7,3)C(5,1) = 35 × 5 = 175 Total 70 + 210 + 175 = 455, which agrees.
2. In how many ways can the five grade-11 members be arranged in a row so that Cara and Dev are not adjacent?
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Count the complement: arrangements with Cara and Dev together.
Treat Cara and Dev as one glued block. The block plus the other three students makes 4 units, arranged in 4! = 24 ways, and the block has 2 internal orders (Cara-Dev or Dev-Cara). So 24 × 2 = 48 arrangements have them adjacent.
Total unrestricted arrangements: 5! = 120.
Not adjacent: 120 - 48 = 72.
Direct check by the gap method: seat the other three first (3! = 6 ways), producing four gaps, and place Cara and Dev in two different gaps in P(4,2) = 12 ways. 6 × 12 = 72, which agrees. The judgement here is to notice that 'not adjacent' is a complement condition, and that the block method must include the internal ordering of the block; forgetting the factor of 2 is the usual source of the wrong answer 96.
3. For each possible committee, a chair and a secretary are chosen from its four members. How many committee-plus-roles outcomes are possible? Explain which counting error a student is likely to make.
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Every 4-member committee supports exactly 4 × 3 = 12 ordered role assignments (chair first, secretary second, distinct people). Since the grade restriction applies to the committee itself and not to the offices:
455 × 12 = 5460 outcomes
Consistency check by subtraction from the unrestricted case: unrestricted role assignments = C(12,4) × 12 = 495 × 12 = 5940. Remove the all-grade-12 committees (35 × 12 = 420) and the all-grade-11 committees (5 × 12 = 60): 5940 - 420 - 60 = 5460. The subtraction is legitimate because each excluded committee carries the same 12 role assignments as any other.
The likely error is to use C(4,2) = 6 for the offices, forgetting that chair and secretary are distinct, which would halve the answer to 2730. A second error is to bolt the roles on before the grade restriction, which produces counts that cannot be corrected afterwards.