Counting Methods - ACT Math

1. How many ways can you choose 2 students from a group of 8?
   Answer: 28 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!.
2. 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?
   Answer: 30 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.
3. A teacher has 7 different problems and will assign 3 of them as a quiz in a specific order (Problem 1, then 2, then 3). Since order matters, how many different quizzes are possible?
   Answer: 210 Since problems are assigned in a specific order (Problem 1, then 2, then 3), this is a permutation problem. We calculate P(7,3) = 7!/(7-3)! = 7!/4! = 7×6×5 = 210. We have 7 choices for the first problem, 6 for the second, and 5 for the third. Choice A (35) incorrectly used the combination formula C(7,3) instead.
4. A school has 10 students and needs to choose 4 of them to attend a workshop. Since only the set of students matters, order does not matter. How many ways can 4 students be chosen from 10?
   Answer: 210 Since only the set of students matters and order does not matter, this is a combination problem. We need to choose 4 students from 10 students without regard to order. Using the combination formula C(10,4) = 10!/(4!×6!) = (10×9×8×7)/(4×3×2×1) = 5040/24 = 210. The calculation divides by 4! to eliminate all possible orderings of the 4 chosen students. Choice B incorrectly used the permutation formula P(10,4) = 5040.
5. A team has 6 players and needs to choose 2 captains (Captain 1 and Captain 2). Since the roles are distinct, order matters. In how many ways can the captains be chosen?
   Answer: 30 Since the roles (Captain 1 and Captain 2) are distinct and order matters, this is a permutation problem. We need to select 2 captains from 6 players where the specific role matters. Using the permutation formula P(6,2) = 6!/(6-2)! = 6!/4! = 6×5 = 30. The calculation shows 6 choices for Captain 1 and 5 remaining choices for Captain 2. Choice A incorrectly used combination C(6,2) = 15, ignoring that captain roles are distinct.
6. How many permutations of 6 distinct items taken 3 at a time are there? (That is, how many ordered arrangements of length 3 can be formed from 6 different items?)
   Answer: 120 Since we want ordered arrangements and order matters, this is a permutation problem. We need to find P(6,3), the number of ways to arrange 3 items from 6 distinct items. Using the permutation formula P(6,3) = 6!/(6-3)! = 6!/3! = 6×5×4 = 120. The calculation shows 6 choices for first position, 5 for second, and 4 for third position. Choice A incorrectly used combination C(6,3) = 20, ignoring that arrangement order matters.
7. How many permutations of 6 distinct items taken 2 at a time are there? (That is, how many ordered pairs can be formed from 6 different items without repetition?)
   Answer: 30 Permutations of 6 items taken 2 at a time means selecting and arranging 2 items where order matters. We calculate P(6,2) = 6!/(6-2)! = 6!/4! = 6×5 = 30. This gives us 6 choices for the first position and 5 for the second position. Choice A (15) represents the combination C(6,2) where order wouldn't matter.
8. How many ways can 4 students be lined up for a group photo?
   Answer: 24 Since we're arranging students in a line, order matters - we use permutations. We need to calculate $4! = 4 \times 3 \times 2 \times 1 = 24$ different arrangements. Each position in the line represents a different arrangement of the same 4 students. Choice C incorrectly calculated $2^3 = 8$ instead of 4!.
9. A group of 7 people is planning a hiking trip. If 3 people must stay behind to prepare dinner, how many ways can the hiking group be chosen?
   Answer: 35 Since we're choosing people for the hiking group (the remaining 4 after 3 stay behind), order doesn't matter - we use combinations. We need C(7,4) = C(7,3) = 7!/(3! × 4!) = (7 × 6 × 5)/(3 × 2 × 1) = 210/6 = 35 ways to choose the hiking group. Choosing 4 to hike is equivalent to choosing 3 to stay behind. Choice B used incorrect calculation.
10. A class has 10 students. The teacher wants to choose 1 student as class president and 1 different student as vice president. Since the positions are different, order matters. In how many ways can the teacher choose the two students?
    Answer: 90 Since the positions are different (president and vice), order matters in assignment, so this is a permutation problem. We use P(10,2) = 10! / (10-2)! = 10 × 9. Calculating: 10 × 9 = 90. This counts distinct ordered pairs without repetition. A key distractor is choice A (45), which is C(10,2), ignoring that positions are distinct.
11. A student has 6 different books (A, B, C, D, E, F) and wants to arrange exactly 4 of them on a shelf in a row. Since the order on the shelf matters, how many possible arrangements are there?
    Answer: 360 Since the order on the shelf matters, this is a permutation problem. We need to find the number of ways to arrange 4 books from 6 different books. Using the permutation formula P(6,4) = 6!/(6-4)! = 6!/2! = 6×5×4×3 = 360. The calculation shows 6 choices for first position, 5 for second, 4 for third, and 3 for fourth position. Choice C incorrectly used 5! = 120, which would be arranging 5 items completely.
12. A pizza shop offers 6 different toppings. A customer chooses exactly 2 different toppings, and the order of toppings does not matter. How many ways can the customer choose the toppings?
    Answer: 15 Since the order of toppings does not matter, this is a combination problem. We need to choose 2 toppings from 6 different toppings without regard to order. Using the combination formula C(6,2) = 6!/(2!×4!) = (6×5)/(2×1) = 30/2 = 15. The calculation divides by 2! to eliminate the ordering of the two chosen toppings. Choice A incorrectly used the permutation formula P(6,2) = 30.
13. A student has 6 different photos and wants to arrange exactly 3 of them in a row in a frame. Since the order in the frame matters, how many possible arrangements are there?
    Answer: 120 Since the order in the frame matters, this is a permutation problem. We need to find the number of ways to arrange 3 photos from 6 different photos. Using the permutation formula P(6,3) = 6!/(6-3)! = 6!/3! = 6×5×4 = 120. The calculation shows 6 choices for first position, 5 for second position, and 4 for third position in the frame. Choice C incorrectly used combination C(6,3) = 20, ignoring that arrangement order matters.
14. In how many ways can 4 friends stand in a line for a photograph?
    Answer: 24 Order matters when friends stand in a line for a photograph, so we use permutation. We need to arrange 4 friends in 4 positions. The calculation is 4! = 4 × 3 × 2 × 1 = 24 arrangements. Each position in the line is distinct, making this a straightforward permutation problem.
15. In how many different ways can a committee of 4 be selected from a group of 7 people?
    Answer: 35 A committee has no ranked positions, so order does not matter and this is a combination, not a permutation. Computing $C(7,4) = \frac{7!}{4! \cdot 3!} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = \frac{210}{6} = 35$ gives the number of possible committees. The answer $840$ is $P(7,4) = 7 \cdot 6 \cdot 5 \cdot 4$, which counts every ordering of the same four people as a different committee, and $210$ is that same product left undivided by only part of the arrangement count, while $49$ is just $7^2$ and has no combinatorial meaning here. Ask whether swapping two chosen people creates a genuinely different outcome: if it does not, divide by the factorial of the group size to strip out the orderings.
16. A club has 8 members and needs to choose 3 people to serve on a committee. Since the committee has no positions, order does not matter. How many ways can the 3 people be chosen?
    Answer: 56 Since committee members have no specific positions and order doesn't matter, this is a combination problem. We calculate C(8,3) = 8!/(3!×5!) = (8×7×6)/(3×2×1) = 336/6 = 56. We divide by 3! to eliminate duplicate arrangements of the same 3 people. Choice B (336) represents the permutation P(8,3) where order would matter.
17. A student rolls a standard 6-sided die and then flips a coin. What is the number of possible outcomes for the ordered pair (die result, coin result)?
    Answer: 12 Since the outcome is an ordered pair (die, coin) and the events are independent, order matters in the sequence, using the multiplication principle. There are 6 outcomes for the die and 2 for the coin, so total is 6 × 2. Calculating: simply 12. This applies the fundamental counting rule for combined experiments. A key distractor is choice C (36), which might come from mistakenly using 6 × 6, as if two dice.
18. A deck of 52 cards is shuffled. How many ways can the top 5 cards be drawn?
    Answer: 311,875,200 Since we're drawing cards in sequence from the top of a deck, order matters - we use permutations. We need P(52,5) = 52!/(52-5)! = 52!/47! = 52 × 51 × 50 × 49 × 48 = 311,875,200 different ways. Each position represents a different card drawn in sequence from the shuffled deck. Choice B incorrectly used the combination formula C(52,5).
19. How many different ways can 5 books be arranged on a shelf?
    Answer: 120 Since we're arranging books on a shelf, order matters - we use permutations. We need to calculate the number of ways to arrange 5 distinct books, which is 5! = 5 × 4 × 3 × 2 × 1 = 120. Each position on the shelf represents a different arrangement of the same books. Choice D incorrectly calculated 4! instead of 5!.
20. A password consists of 1 letter followed by 2 digits. The letter can be chosen from 5 options (A-E), and each digit can be chosen from 0-9. Digits may repeat. What is the number of possible passwords?
    Answer: 500 This is a multiplication principle problem with independent choices for each position. We have 5 choices for the letter, 10 choices for the first digit, and 10 choices for the second digit (since digits may repeat). Total passwords = 5 × 10 × 10 = 500. Choice A (150) might result from incorrectly not allowing digit repetition.

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