Quiz Team Selection with Conditional Rules and a Counterintuitive Claim
An inter-college quiz team of exactly five students must be chosen from eight: A, B, C, D from Science and E, F, G, H from Arts. The selection rules are: (i) at least two Science and at least two Arts students must be on the team; (ii) if A is selected, B cannot be; (iii) if C is selected, E must be; (iv) if F is selected, G cannot be; (v) if D is selected, H must be. The student union proposes the team {A, C, E, F, H}. A second proposal is {B, C, E, G, H}. The faculty coordinator wants each proposal checked rule by rule, not by impression. One student argues: 'Any valid team that contains A must also contain E, because A and E are the strongest speakers.' Another suggests replacing rule (iv) with: 'F and G must be selected together or neither selected.'
1. Check each proposed team against all five rules. State which, if either, is valid.
Read the worked answer
Team 1: {A, C, E, F, H}
- Science count: A and C = 2; Arts count: E, F, H = 3. Rule (i) satisfied.
- A is in and B is out: rule (ii) satisfied.
- C is in and E is in: rule (iii) satisfied.
- F is in and G is out: rule (iv) satisfied.
- D is out, so rule (v) imposes nothing. Team 1 is valid.
Team 2: {B, C, E, G, H}
- Science count: B and C = 2; Arts count: E, G, H = 3. Rule (i) satisfied.
- A is out, so rule (ii) imposes nothing.
- C is in and E is in: rule (iii) satisfied.
- F is out, so rule (iv) imposes nothing.
- D is out, so rule (v) imposes nothing. Team 2 is also valid. Both proposals satisfy every stated rule.
2. A student claims: 'Any valid team containing A must also contain E.' Test the claim logically. If it is true, prove it; if false, give a counterexample.
Read the worked answer
The claim is true, but not for the reason the student gives. The proof uses the constraints.
Assume A is selected. By rule (ii), B is excluded. Suppose, for contradiction, that E is not selected. Then rule (iii) excludes C, because selecting C would require E. The only students still available are A, D, F, G, H. Rule (iv) says F and G cannot both be selected, so from {D, F, G, H} we can take at most three students: D, H, and at most one of F or G. Adding A gives at most (1+3=4) students. But the team must have exactly five. Contradiction. Therefore E must be selected whenever A is selected.
The student's stated reason ('strongest speakers') is irrelevant; the necessity comes from the rule structure. A correct explanation should mention the exclusion of B, the exclusion of C when E is absent, and the F/G conflict.
3. Replace rule (iv) with 'F and G must be selected together or neither.' Does the first proposed team remain valid? Find one valid five-person team under the revised rules that contains both F and G.
Read the worked answer
The first team {A, C, E, F, H} contains F but not G. Under the revised rule, selecting F forces G, so the team is invalid. The new rule also means selecting G would force F.
One valid team is {A, C, E, F, G}:
- Science: A and C = 2; Arts: E, F, G = 3, so rule (i) holds.
- A is in, B is out: rule (ii) holds.
- C is in, E is in: rule (iii) holds.
- F and G are both in: revised rule (iv) holds.
- D is out: rule (v) holds.
So {A, C, E, F, G} satisfies the revised set. Notice that the replacement makes the earlier proof about A forcing E no longer automatically transferable; each rule change must be rechecked.