Order matters when arranging letters to form different words, so we use permutation. The word 'CAT' has 3 distinct letters to arrange in 3 positions. The calculation is 3! = 3 × 2 × 1 = 6 arrangements. These would be: CAT, CTA, ACT, ATC, TCA, TAC.

Since we're arranging books in a row where left-to-right order matters, this is a permutation problem. We need to find P(6,4) = 6!/(6-4)! = 6!/2! = 6×5×4×3 = 360. The calculation gives us 6 choices for the first position, 5 for the second, 4 for the third, and 3 for the fourth position. Choice A (15) incorrectly used the combination formula C(6,4) instead of permutation.

Since we're choosing different officers (president, vice-president, secretary), order matters - we use permutations. We need P(8,3) = 8!/(8-3)! = 8!/5! = 8 × 7 × 6 = 336 ways to fill the three distinct positions. Each office represents a different role among the 8 members. Choice B incorrectly used the combination formula C(8,3) = 56.

Order does not matter when choosing students for a group, so we use combination. We need to choose 2 students from 8 available students. The calculation is C(8,2) = 8!/(2! × 6!) = (8 × 7)/(2 × 1) = 56/2 = 28. Choice B incorrectly used 8 × 7 = 56 without dividing by 2!.

For circular arrangements, we fix one person to account for rotational symmetry and arrange the rest. With 5 people, we fix one person and arrange the remaining 4, giving us (5-1)! = 4! = 4 × 3 × 2 × 1 = 24 arrangements. Circular permutations eliminate rotational duplicates that would occur in linear arrangements. Choice B incorrectly used 5! = 120 for linear arrangement.

Since we're arranging 5 single matches in sequence, order matters - we use permutations. We need 5! = 5 × 4 × 3 × 2 × 1 = 120 different ways to arrange the matches in the tournament schedule. Each position represents a different time slot for the distinct matches. Choice B incorrectly calculated 5!/2! = 60.

Since we're choosing a subcommittee, order doesn't matter - we use combinations. We need to calculate C(6,3) = 6!/(3! × 3!) = (6 × 5 × 4)/(3 × 2 × 1) = 120/6 = 20. This represents selecting 3 members from 6 friends where the arrangement doesn't affect the subcommittee composition. Choice B incorrectly used the permutation formula P(6,3) = 120.

Since we're awarding different medals (gold, silver, bronze), order matters - we use permutations. We need P(4,3) = 4!/(4-3)! = 4!/1! = 4 × 3 × 2 = 24 ways to award the three different medals. Each medal represents a distinct ranking among the 4 participants. Choice C incorrectly used the combination formula C(4,3) = 4.

Since we're distributing different prizes to specific participants, order matters - we use permutations. We need P(5,3) = 5!/(5-3)! = 5!/2! = 5 × 4 × 3 = 60 ways to distribute the three distinct prizes. Each participant can receive at most one prize, making this a permutation problem. Choice B incorrectly used the combination formula C(5,3) = 10.

Since the snacks are chosen as an unordered pair but the drink is separate, the snacks use combination while the drink uses multiplication principle. For snacks, C(5,2) = 5! / (2! × 3!) = (5 × 4) / 2 = 10; then multiply by 3 drinks. Total: 3 × 10 = 30. This combines combination for unordered part with direct counting. A key distractor is choice D (60), which uses P(5,2) = 20 for snacks instead, assuming order matters for snacks.