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ACT Counting & combinations practice questions
Counting & combinations sits in the Statistics & Probability reporting category on the ACT Math section. Prep36 has 10 questions tagged to this skill, every one with a worked explanation. Three of them are below in full, with the reasoning, so you can see what the skill actually looks like before you practise it.
What Counting & combinations covers on the ACT
Within this skill the Prep36 bank covers fundamental counting principle, combinations of a set, permutations for ordered awards, combinations with category constraints, permutations and combinations with at-least cases.
Math is 45 questions in 50 minutes, which is about 66 seconds each, and it is one of the three sections that average into your composite. The difficulty split across the 10 questions tagged here is 1 easier, 6 medium and 3 harder.
Questions are original, written to the Enhanced ACT format. Prep36 is not affiliated with ACT, Inc.
3 counting & combinations questions, with the reasoning
Question 1 easy
At a ring-toss prize counter, a winner picks 1 of 4 stuffed animals, 1 of 3 colors, and 1 of 2 sizes. How many different prize combinations are possible?
- A. 9
- B. 12
- C. 18
- D. 24 correct
Why: By the fundamental counting principle, multiply the number of independent choices: 4 × 3 × 2 = 24, choice D. The choice 9 adds the options as 4 + 3 + 2 instead of multiplying. The choice 12 multiplies only the animal and color and forgets the size. The choice 18 mismultiplies, for example 3 × 3 × 2. Tip: independent choices multiply, they do not add. When each stage is a separate free pick, string the counts together with multiplication signs.
Question 2 medium
A robotics club has 8 members. In how many different ways can the club choose a president, a vice president, and a treasurer, if no member can hold more than one office?
- A. 24
- B. 56
- C. 336 correct
- D. 512
Why: The three offices are distinct, so order matters and this is a permutation: 8 choices for president, 7 left for vice president, 6 left for treasurer, giving 8 × 7 × 6 = 336. The distractors each break one condition: 56 is the combination C(8,3), which counts unordered groups and ignores who gets which title, 512 is 8³, which lets one person hold all three offices, and 24 is just 8 × 3. Tip: ask whether swapping two selected people creates a different outcome. If yes, it is a permutation; if no, it is a combination.
Question 3 hard
A club has 5 women and 4 men. A committee of 4 people will be formed that must contain exactly 2 women and exactly 2 men. How many different committees are possible?
- A. 16
- B. 40
- C. 60 correct
- D. 126
Why: Pick the women in C(5,2) = 10 ways and the men in C(4,2) = 6 ways. Those two picks happen together on the same committee, so multiply: 10 × 6 = 60. The choice 126 is C(9,4), every 4-person committee from all 9 members, which ignores the exactly-2-and-2 requirement. The choice 16 adds 10 + 6 instead of multiplying, and 40 pairs the wrong two counts. Tip: separate choices that all have to happen get multiplied. Choices that are alternatives to each other get added. Decide which one you are looking at before you touch the numbers.
The explanation is the point. Getting a question wrong tells you nothing you did not already suspect; knowing which step you missed is what changes the next one.
Practise the rest of the 10
The other 7 questions on this skill are in the free question bank, filterable by difficulty, each with the same worked explanation. No account and no card.
Open the Math bank
Frequently asked questions
How many ACT counting & combinations questions does Prep36 have?
10, each tagged to this skill and each with a worked explanation.
Are these real ACT questions?
No. They are original questions written to the Enhanced ACT format. Prep36 is not affiliated with ACT, Inc.
Is the Math section part of the composite?
Yes. On the Enhanced ACT the composite is the average of English, Math and Reading. Science is optional and reported separately.
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