Constituting the Five-Member Special Inspection Team
An ESIC zonal office must send a five-member special inspection team to audit a cluster of nine factories over five working days. Seven officers - P, Q, R, S, T and V, together with U - are available, and the zonal rules for constituting such teams are these: (1) if P is included, Q must be included; (2) R and S cannot both be included, since they are required to sign each other's verification sheets and one must stay behind; (3) if T is included, U cannot be included, because both are needed at the zonal office on the same days; (4) at least one of U and V must be included, since only they hold the biometric access cards. The team must have exactly five members and any team that violates a rule is sent back for revision. The office assistant drafts the team {P, Q, S, U, V} and sends it for approval. The Deputy Director wants to know whether the rules leave any room for choice, whether S can ever be included, and whether Q and R are unavoidable.
1. How many valid five-member teams exist? List them.
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Start with rule 4. Suppose V is out; then U is in and rule 3 bars T. With V and T both out, the only candidates left are P, Q, R, S and U - exactly five - so all of them must go, but that puts R and S together in breach of rule 2. So V cannot be out: V is in every valid team. Now split on U.
U in: T is out, and three of P, Q, R, S must join U and V. Dropping S gives {P, Q, R, U, V}, which satisfies rules 1 and 5. Dropping R leaves Q without R; dropping Q leaves P without Q; dropping P leaves R with S. So exactly one team.
U out: V and T may both be in, and four of P, Q, R, S, T must be chosen. Dropping S gives {P, Q, R, T, V}; every other omission breaks rule 1, the Q-needs-R link, or rule 2. So exactly one team.
Two valid teams: {P, Q, R, U, V} and {P, Q, R, T, V}. The draft {P, Q, S, U, V} is rejected because Q appears without R.
2. Is the claim 'S can never be on a valid team' correct?
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It is correct, and the proof is short. Assume S is in. By rule 2, R is out. By the Q-needs-R link (rule 1 read with Q requiring R), Q is out, and with Q out, P is out, since P cannot be included without Q. That removes P, Q and R, leaving only S, T, U and V - four officers for a team of five. There is no fifth person to add, so no valid team can contain S. Notice that this conclusion does not depend on rule 4 at all; it follows from rules 1 and 2 plus the fixed team size. The practical value of the check is that it tells the office not to waste time on drafts containing S: any such draft is dead on arrival, and the assistant's proposal could have been rejected without examining the U and V question.
3. A note says every valid team must contain both Q and R. Verify.
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The claim is true, and each half needs its own argument. R is unavoidable: if R is out, Q is out too, because Q cannot be included without R. Removing both leaves P, S, T, U, V - exactly five for a five-member team - so all must be taken, but T and U cannot sit together under rule 3. Contradiction, so R is in. Q is unavoidable: if Q is out, P is out as well, since P cannot come without Q. The remaining pool is R, S, T, U, V - again exactly five - so all must be taken, but R and S cannot coexist under rule 2. Contradiction, so Q is in. Both valid teams, {P, Q, R, U, V} and {P, Q, R, T, V}, bear this out. So the Deputy Director can be told that the composition is fixed on three names - P, Q, R - and the only genuine discretion lies between U and T.