ACT Math Quiz: Counting Methods

What this quiz covers

This quiz focuses on Counting Methods, giving you a quick way to practice the rules, question types, and explanations that matter most for ACT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 12
2. 36
3. 15 (correct answer)
4. 30

Explanation: Because only which questions are chosen matters and order does not, this is a combination, counted by $C(n,r)=\dfrac{n!}{r!(n-r)!}$. With 6 questions taken 2 at a time, $C(6,2)=\dfrac{6!}{2!\,4!}=\dfrac{6\times5}{2\times1}=\dfrac{30}{2}=15$, so there are 15 possible selections; dividing by $2!$ removes the double counting of each pair listed in two orders. The value 30 is the permutation $P(6,2)=6\times5$, which is what you get by skipping that division and treating the two picks as ordered. The value 36 comes from computing $6^2$, as if each of two slots could independently be any of the 6 questions, and 12 comes from multiplying $6\times2$ instead of using a counting formula at all. Decide whether order matters before you compute, since ordered arrangements use $n!/(n-r)!$ while unordered selections divide that result by $r!$.

Question 2

1. 15
2. 24
3. 360 (correct answer)
4. 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

1. 24
2. 56
3. 512
4. 336 (correct answer)

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

1. 8
2. 56
3. 16
4. 28 (correct answer)

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

1. 120 (correct answer)
2. 60
3. 24
4. 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 6

1. 45
2. 75
3. 30 (correct answer)
4. 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.

Question 7

1. 15
2. 24
3. 120
4. 360 (correct answer)

Explanation: Since the left-to-right order on the shelf matters, this is a permutation problem where we arrange 4 books from 6 available books. We use the permutation formula P(6,4) = 6!/(6-4)! = 6!/2! = 6×5×4×3 = 360. The calculation proceeds as 6×5 = 30, then 30×4 = 120, and finally 120×3 = 360. Choice C (120) represents the error of stopping the calculation too early at 6×5×4 without including the final multiplication by 3.

Question 8

1. 9
2. 20 (correct answer)
3. 45
4. 10

Explanation: This is a multiplication principle problem where we make two independent choices. We have 4 choices for the drink and 5 choices for the snack, and any drink can pair with any snack. Total pairs = 4 × 5 = 20. Choice C (45) might result from incorrectly adding the options (4+5) and then choosing 2.

Question 9

1. 56 (correct answer)
2. 336
3. 112
4. 24

Explanation: Since the committee roles are identical and order does not matter, this is a combination problem. We use the combination formula C(n,k) = n! / (k! (n-k)! ) to choose 3 members from 8. Here, C(8,3) = 8! / (3! × 5!) = (8 × 7 × 6 × 5!) / (6 × 5!) = (8 × 7 × 6) / 6 = 56. Simplifying: 8 × 7 = 56, then divided by 2 (from 3!/3, but correctly it's 336 / 6 = 56). A key distractor is choice B (336), which uses the permutation P(8,3) instead, incorrectly assuming order matters.

Question 10

1. 10 (correct answer)
2. 20
3. 5
4. 15

Explanation: Order does not matter when choosing chefs for a task, so we use combination. We need to choose 2 chefs from 5 available chefs. The calculation is C(5,2) = 5!/(2! × 3!) = (5 × 4)/(2 × 1) = 20/2 = 10. This gives all possible pairs of chefs who could work together.

Question 11

1. 5
2. 24
3. 60
4. 120 (correct answer)

Explanation: Order matters when arranging beads on a string, so we use permutation. We need to arrange 5 different colored beads in 5 positions. The calculation is 5! = 5 × 4 × 3 × 2 × 1 = 120 arrangements. Each bead occupies a distinct position on the string, making this a straightforward permutation.

Question 12

1. 49
2. 210
3. 840
4. 35 (correct answer)

Explanation: 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.

Question 13

1. 3
2. 6 (correct answer)
3. 12
4. 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 14

1. 24
2. 56 (correct answer)
3. 336
4. 512

Explanation: This is a combinations question testing when to use C(n, r) versus P(n, r). Choice B (56) is correct — order does not matter for forming a panel (a group of 3 students, not a ranked list), so use combinations: ₈C₃ = 8! / (3! × 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56. Choice A (24) results from computing 3! × 4 = 24 or another partial calculation that underestimates the number of combinations. Choice C (336) uses the permutation formula ₈P₃ = 8 × 7 × 6 = 336 instead of combinations — counting ordered arrangements rather than unordered groups. This overcounts by a factor of 3! = 6. Choice D (512) computes 8^(something) or 2⁹, a miscalculation. Pro tip: Ask yourself: does the order of selection matter? For committees, teams, and panels — no. Use combinations C(n, r). For rankings, sequences, or assigned roles — yes. Use permutations P(n, r). The difference is always dividing by r! to remove ordering.

Question 15

1. 20
2. 40 (correct answer)
3. 80
4. 84

Explanation: The correct answer is B (40). Use combinations: choose 2 seniors from 5, then 1 junior from 4. C(5, 2) × C(4, 1) = [5!/(2!3!)] × 4 = 10 × 4 = 40. A (20) comes from choosing 1 senior and 1 junior: C(5,1) × C(4,1) = 5 × 4 = 20 — uses the wrong number of seniors. C (80) doubles the correct answer, possibly adding an incorrect multiplier. D (84) computes C(9, 3) = 84 — the total number of committees from all 9 students without applying the constraint of exactly 2 seniors and 1 junior. Always apply combinations separately for each constrained group, then multiply.

Question 16

1. 26
2. 17,576
3. 15,600 (correct answer)
4. 6,240

Explanation: Since the password consists of different letters in specific positions, order matters - we use permutations. We need P(26,3) = 26!/(26-3)! = 26!/23! = 26 × 25 × 24 = 15,600 different passwords. Each position in the password represents a distinct choice from the remaining letters. Choice B incorrectly allowed repetition with 26³ = 17,576.

Question 17

1. 15
2. 20
3. 10 (correct answer)
4. 30

Explanation: Since we have identical balls of each color, we use combinations for arrangements with repetition. We need to arrange 5 positions with 3 red and 2 blue balls, which is C(5,3) = C(5,2) = 5!/(3! × 2!) = (5 × 4)/(2 × 1) = 10 different arrangements. This counts the distinct patterns of red and blue positions. Choice B incorrectly used permutations.

Question 18

1. 15
2. 24
3. 120
4. 360 (correct answer)

Explanation: 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.

Question 19

1. 72
2. 3024
3. 36
4. 126 (correct answer)

Explanation: Since the order of selecting players does not matter, this is a combination problem. We use C(9,4) = 9! / (4! × 5!) to choose 4 from 9. Simplifying: (9 × 8 × 7 × 6 × 5!) / (24 × 5!) = (9 × 8 × 7 × 6) / 24 = 3,024 / 24 = 126. Step by step: 9/1 × 8/2 × 7/3 × 6/4 = 9 × 4 × (7/3) × (3/2), but directly divides to 126. A key distractor is choice B (3024), which is the numerator without dividing by 4!, confusing with permutations.

Question 20

1. 30
2. 15 (correct answer)
3. 12
4. 36

Explanation: Since the order of toppings doesn't matter (pepperoni-mushroom is the same as mushroom-pepperoni), this is a combination problem. We calculate C(6,2) = 6!/(2!×4!) = (6×5)/(2×1) = 30/2 = 15. We choose 2 toppings from 6 available options. Choice A (30) represents the permutation P(6,2) where order would matter.