Question 1
In how many ways can you arrange the letters in the word 'CAT'?
- 3
- 6
- 12
- 9
Explanation: 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.
Question 2
A student has 6 different books (A, B, C, D, E, F) and wants to arrange exactly 4 of them in a row on a shelf. Since the left-to-right order matters, how many possible arrangements are there?
- 15
- 24
- 360
- 120
Explanation: 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.
Question 3
A club has 8 members. How many different ways can the club choose a president, vice-president, and secretary?
- 24
- 56
- 512
- 336
Explanation: 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.
Question 4
How many ways can you choose 2 students from a group of 8?
- 8
- 56
- 16
- 28
Explanation: 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!.
Question 5
In how many ways can 5 people be seated around a circular table?
- 24
- 120
- 60
- 30
Explanation: 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.
Question 6
In a tournament, there are 5 single matches. How many ways can the matches be arranged if no match can be repeated?
- 120
- 60
- 24
- 30
Explanation: 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.
Question 7
A group of 6 friends is trying to form a subcommittee of 3 members. How many ways can they choose the subcommittee members?
- 20
- 120
- 60
- 18
Explanation: 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.
Question 8
In a race with 4 participants, how many ways can the gold, silver, and bronze medals be awarded?
- 24
- 12
- 6
- 10
Explanation: 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.
Question 9
How many ways can 3 different prizes be distributed among 5 participants if each participant can receive only one prize?
- 60
- 10
- 120
- 20
Explanation: 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.
Question 10
A snack pack is made by choosing 1 drink from 3 options and 2 different snacks from 5 options. The two snacks are chosen as a pair (order does not matter), but the drink choice is separate. What is the number of possible snack packs?
- 45
- 75
- 30
- 60
Explanation: 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.