Selecting and seating a school disciplinary committee with restrictions
At a higher secondary school in Chennai, the principal must constitute a five-member disciplinary committee from 12 prefects: 7 from the science stream and 5 from the commerce stream. School rules require at least two members from each stream. Two prefects, A (science) and B (commerce), have had a public disagreement, and the principal has decided they may not both serve on the same committee. A parent-teacher association member suggests choosing any five prefects because all prefects are equally responsible, but the principal wants a count that respects both the stream rule and the A-B restriction before placing the proposal before the school management committee. The senior commerce prefect C is expected to be in the final committee. Once the committee is fixed, its five members must be seated around a circular table for a group photograph. The approved committee has 3 science and 2 commerce prefects, includes A but not B, and includes C among the commerce members.
1. Ignoring the A-B restriction for the moment, how many five-member committees satisfy the rule of at least two members from each stream?
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A five-member committee must have either 2 science and 3 commerce members or 3 science and 2 commerce members; a 1-4 split is not allowed.
Case 2 science, 3 commerce: choose 2 of 7 science prefects and 3 of 5 commerce prefects, giving C(7,2)C(5,3) = 21 × 10 = 210.
Case 3 science, 2 commerce: C(7,3)C(5,2) = 35 × 10 = 350.
These cases are disjoint, so the total is 210 + 350 = 560 committees. The parent-teacher association suggestion would give C(12,5) = 792 committees, which ignores the stream rule.
2. Now apply the restriction that A and B cannot both serve. How many committees satisfy both the stream rule and the A-B restriction?
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From the 560 committees, subtract those containing both A and B.
If A and B are both included, one science and one commerce place is already filled, so 3 more members must be chosen from the remaining 6 science and 4 commerce prefects. The stream rule still requires at least one more from each stream among these 3 members. The possible additions are:
1 science and 2 commerce: C(6,1)C(4,2) = 6 × 6 = 36. 2 science and 1 commerce: C(6,2)C(4,1) = 15 × 4 = 60.
So 36 + 60 = 96 committees contain both A and B.
Valid committees = 560 − 96 = 464. The subtraction is legitimate because every forbidden committee counted in the 96 also satisfies the stream rule; no committee is subtracted twice.
3. The approved five-member committee has 3 science and 2 commerce prefects, includes A but excludes B, and includes C among the commerce members. In how many ways can the five members be seated around a circular table if A must not sit next to C?
Read the worked answer
First count all circular arrangements of the 5 distinct members: (5 − 1)! = 24.
Now count the arrangements in which A and C are adjacent. Treat the pair A-C as one block. The block plus the other 3 members gives 4 objects around a circle, arranged in (4 − 1)! = 6 ways. Within the block, A and C can be ordered in 2! = 2 ways. So adjacent arrangements = 6 × 2 = 12.
Valid arrangements = 24 − 12 = 12. Equivalently, seat the other three members first in (3 − 1)! = 2 circular orders; there are then 3 gaps between them, but A and C must occupy two non-adjacent gaps, which can be chosen in C(3,2) = 3 ways, and A and C can be ordered in 2 ways: 2 × 3 × 2 = 12.