ACT · Math · Preparing for Higher Math: Statistics and Probability

Counting and combinatorics

Counting and combinatorics is part of Preparing for Higher Math: Statistics and Probability in the ACT Math section. The questions below are written in the style of the real test, each with the answer and a worked explanation.

Question types you will see here include multiplication principle, arrangements of distinct items, selection where order does not matter and arrangements with repeated items.

Common traps: adding the choices; multiplying only some of the stages; counting a choice that is not allowed.

Sample questions

Question 1
Six friends sit around a round table. Seatings that differ only by rotating everyone around the table are considered the same. How many different seatings are possible?
- A.
  6
- B.
  60
- C.
  120
- D.
  720
Answer: C. 120
Fix one friend's seat to remove rotations; the other 5 can be arranged in 5! = 120 ways. Choice A is just the number of people; B also divides by 2 for mirror images, which the question does not do; D treats rotations as different seatings.
Question 2
In a race with 9 runners, how many different ways can 1st through 4th place be awarded, assuming no ties?
- A.
  36
- B.
  126
- C.
  3024
- D.
  6561
Answer: C. 3024
Order matters, so use permutations: 9 × 8 × 7 × 6 = 3024.
Question 3
A playlist has n different songs. There are exactly 132 ways to choose a first song and a different second song to play, in order. What is the value of n?
- A.
  11
- B.
  12
- C.
  66
- D.
  132
Answer: B. 12
Ordered choices of 2 from n number n(n − 1). Since 12 × 11 = 132, n = 12. Choice A is the second factor, n − 1; C halves 132 as if order did not matter, and gives a count, not n; D confuses the number of ordered pairs with the number of songs.
Question 4
At a conference in Singapore, each of 9 tutors exchanges one business card with every other tutor, exactly once per pair. How many exchanges take place?
- A.
  17
- B.
  18
- C.
  36
- D.
  72
Answer: C. 36
Each pair swaps once, so the count is C(9, 2) = 9 × 8 ÷ 2 = 36. Choice A adds 9 + 8; B multiplies 9 × 2; D is 9 × 8, which counts each pair twice.
Question 5
A 4-digit door code uses four different digits. The first digit cannot be 0, and the last digit must be odd. How many codes are possible?
- A.
  2,240
- B.
  2,520
- C.
  4,536
- D.
  5,040
Answer: A. 2,240
Fill the restricted places first. Last digit: 5 odd choices. First digit: not 0 and not the last digit, 8 choices. Middle digits: 8 × 7. Total 5 × 8 × 8 × 7 = 2,240. Choice B ignores the no-zero rule for the first digit; C ignores the odd rule (9 × 9 × 8 × 7); D ignores both rules (10 × 9 × 8 × 7).

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