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ACT Math Help: Counting Methods

Review real example questions for Counting Methods in ACT Math.

Question 1 / 10

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A student must answer exactly 2 questions from a set of 6 questions on a quiz. Since only which questions are chosen matters, order does not matter. How many ways can the student choose the questions?

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Question 1

A student must answer exactly 2 questions from a set of 6 questions on a quiz. Since only which questions are chosen matters, order does not matter. How many ways can the student choose the questions?

  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)=n!r!(n−r)!C(n,r)=\dfrac{n!}{r!(n-r)!}. With 6 questions taken 2 at a time, C(6,2)=6!2! 4!=6×52×1=302=15C(6,2)=\dfrac{6!}{2!\,4!}=\dfrac{6\times5}{2\times1}=\dfrac{30}{2}=15, so there are 15 possible selections; dividing by 2!2! removes the double counting of each pair listed in two orders. The value 30 is the permutation P(6,2)=6×5P(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 626^2, as if each of two slots could independently be any of the 6 questions, and 12 comes from multiplying 6×26\times2 instead of using a counting formula at all. Decide whether order matters before you compute, since ordered arrangements use n!/(n−r)!n!/(n-r)! while unordered selections divide that result by r!r!.

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?

  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

A club has 8 members. How many different ways can the club choose a president, vice-president, and secretary?

  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

How many ways can you choose 2 students from a group of 8?

  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

In a tournament, there are 5 single matches. How many ways can the matches be arranged if no match can be repeated?

  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

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?

  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

A code consists of 3 different letters chosen from 6 distinct letters (A-F), and the letters are arranged in order. No letter may repeat. What is the number of possible codes?

  1. 20
  2. 120 (correct answer)
  3. 216
  4. 60

Explanation: Since letters are arranged in order and no letter may repeat, this is a permutation problem. We calculate P(6,3) = 6!/(6-3)! = 6!/3! = 6×5×4 = 120. We have 6 choices for the first letter, 5 for the second (can't repeat), and 4 for the third. Choice C (216) would result from allowing repetition (6×6×6).

Question 8

A student has 6 different books and wants to place exactly 4 of them on a shelf in a row. Since the left-to-right order on the shelf matters, how many possible arrangements of 4 books chosen from the 6 are there?

  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 9

A student must choose 1 drink and 1 snack. There are 4 drink options and 5 snack options. If any drink can be paired with any snack, what is the number of possible pairs?

  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 10

A club has 8 members. The club needs to choose 3 members to serve on a committee, and the roles are identical (so order does not matter). How many different committees are possible?

  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.