The Five-Proposal Funding Puzzle
A student government committee must select exactly three of five proposals - P, Q, R, S, and T - for funding. The committee's bylaws impose four rules. First, P and Q cannot both be funded. Second, R can be funded only if S is also funded. Third, T can be funded only if Q is funded. Fourth, S and T cannot both be funded. The committee has already received written support for all five proposals, but it cannot fund more than three because the total budget is capped at €18,000 and each proposal costs €6,000. A member argues that because T has strong support, the committee should fund T, Q, and one other proposal. Another member warns that this combination may violate the bylaws. The chair asks for a complete list of permissible three-proposal combinations before the vote.
1. List all permissible three-proposal combinations and explain your reasoning.
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T cannot be funded. If T is selected, rule 3 requires Q; rule 1 then excludes P, and rule 4 excludes S. Rule 2 says R requires S, so R is also excluded. That leaves only T and Q, but exactly three proposals must be funded. So any valid set excludes T. With T excluded, the committee must choose three of P, Q, R, S. Rules 1 and 2 imply that exactly one of P and Q is chosen, and that R and S are both chosen. Hence the complete list is {P, R, S} and {Q, R, S}.
2. A member claims: 'If we fund T, we must fund Q; therefore if we fund Q, we must fund T.' Is this inference valid?
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The inference is invalid. The original rule is 'if T, then Q', which makes Q necessary for T. The member's conclusion is 'if Q, then T', which treats a necessary condition as sufficient. The valid set {Q, R, S} is a counterexample: Q is funded, T is not, and no rule is broken. In ordinary reasoning this is the fallacy of affirming the consequent. The practical lesson is that strong support for T cannot be used to infer that T must be included whenever Q is included; the bylaws may make T impossible altogether, as the previous answer shows.
3. Suppose the budget rises to €24,000 and exactly four proposals can be funded. Which combinations are possible under the same bylaws?
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No valid four-proposal combination exists. If T were included, the same chain as before would force Q in, exclude P and S, and exclude R because R requires S; only two proposals would remain. So T is out. From P, Q, R, S, rule 1 prevents choosing both P and Q, so the maximum is three: either P, R, S or Q, R, S. Raising the budget to €24,000 removes the financial cap but does not change the bylaws. The chair should be told that the constraint is legal, not budgetary, and that funding four would require amending at least one rule, for example by allowing P and Q together or by relaxing R's dependency on S.