Study Counting Methods in ACT Math with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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Flashcard 1: How many ways can 5 different keys be arranged on a keyring?
Answer:
- Circular arrangement with identical rotations.
Flashcard 2: What is the number of distinct permutations of the letters in LEVEL?
Answer: 2!2!5!=30. Two L's and two E's create identical arrangements.
Flashcard 3: How many permutations are possible for arranging n distinct objects?
Answer: n!. Each position has decreasing choices, resulting in n! total arrangements.
Flashcard 4: State the number of ways to assign 3 tasks to 3 workers, one task per worker.
Answer:
- One-to-one assignment gives 3!=6 possible arrangements.
Flashcard 5: What is the number of binary strings of length 5 with exactly 2 ones?
Answer: (25)=10. Choose positions for ones among five spots.
Flashcard 6: What is the number of ways to choose r items from n distinct items when order does not matter?
Answer: nCr. Selection without regard to arrangement order.
Flashcard 7: What is the number of permutations where 3 items are identical and 2 are distinct?
Answer: 3!5!. Divides by 3! to account for identical items in permutations.
Flashcard 8: State the number of ways to arrange 3 books on a shelf.
Answer:
- Three distinct books can be arranged in 3!=6 ways.
Flashcard 9: How many ways can you distribute 7 different books to 7 people?
Answer:
- Each book goes to exactly one person.
Flashcard 10: What is the formula for circular permutations of n distinct objects?
Answer: (n−1)!. Fix one object to break circular symmetry.
Flashcard 11: What does n! represent in counting methods?
Answer: The product of all positive integers up to n. Calculated as n×(n−1)×...×2×1.
Flashcard 12: Identify the number of ways to arrange 4 people in a line.
Answer:
- Calculate 4!=4×3×2×1=24 arrangements.
Flashcard 13: What is the number of ways to choose a 2-person committee from 6 people?
Answer: 6C2=15. Committee selection ignores member ordering.
Flashcard 14: How many distinct permutations are there of the letters in 'MISSISSIPPI'?
Answer: 4!4!2!11!=34,650. Divide by repeated letters: 4 I's, 4 S's, 2 P's.
Flashcard 15: What is the number of subsets of a set with 6 elements?
Answer: 26=64. Each element independently included or excluded.
Flashcard 16: Identify the rule: when choices are mutually exclusive, how do you count total outcomes?
Answer: Add the counts: m+n+⋯. Addition principle for disjoint events.
Flashcard 17: How many ways can 8 people sit in a circle?
Answer:
- Fix one position to eliminate rotational symmetry.
Flashcard 18: What is the two-set inclusion–exclusion formula for ∣A∪B∣?
Answer: ∣A∪B∣=∣A∣+∣B∣−∣A∩B∣. Add sets, subtract overlap to avoid double-counting.
Flashcard 19: What is the value of 0!?
Answer:
- By definition, factorial of zero equals one.
Flashcard 20: How many ways can you choose 4 items from 10?
Answer: (410)=210. Combination formula with n=10 and r=4.
Flashcard 21: What is the formula for calculating combinations of n items taken r at a time?
Answer: C(n,r)=r!(n−r)!n!. Order doesn't matter, so divide by r! arrangements.
Flashcard 22: How many ways can you select 3 desserts from a menu of 5?
Answer: (35)=10. Order doesn't matter for dessert selection.
Flashcard 23: How many ways can you arrange 3 red balls and 2 blue balls in a row?
Answer: 3!2!5!=10. Divide by repetitions of identical colored balls.
Flashcard 24: Identify the number of ways to arrange 4 people in a line.
Answer:
- Calculate 4!=4×3×2×1=24 arrangements.
Flashcard 25: How many ways can a committee of 2 be formed from 5 people?
Answer: (25)=10. Order doesn't matter for committee selection.
Flashcard 26: What is the number of distinct permutations of n objects with repeats n1,n2,…,nk?
Answer: n1!n2!⋯nk!n!. Divide by factorials of identical object counts.
Flashcard 27: What is ∣A∪B∣ if ∣A∣=12, ∣B∣=9, and ∣A∩B∣=5?
Answer: 16. Using inclusion-exclusion: 12+9−5=16
Flashcard 28: How many ways can 4 identical items be distributed into 3 different boxes?
Answer:
- Stars and bars: C(4+3−1,3−1)=C(6,2)=15 distributions.
Flashcard 29: How many ways can you distribute 7 different books to 7 people?
Answer:
- Each book goes to exactly one person.
Flashcard 30: What is 6C3?
Answer: 20. 3!3!6!=36720=20
Flashcard 31: What is the relationship between permutations and combinations: nPr in terms of nCr?
Answer: nPr=nCr⋅r!. Permutations equal combinations times internal arrangements.
Flashcard 32: What is a permutation?
Answer: An arrangement of items in a specific order. Order is important in arrangements.
Flashcard 33: What is the formula for calculating permutations of n items taken r at a time?
Answer: P(n,r)=(n−r)!n!. Order matters, so divide by unused positions (n−r)!.
Flashcard 34: What is the principle of multiplication in counting?
Answer: If one event has m outcomes and another has n, total is m×n. Sequential events multiply their outcome counts.
Flashcard 35: What is the formula for permutations with repetition?
Answer: nr. Each position can be filled with any of n items.
Flashcard 36: What is the number of binary strings of length 5 with exactly 2 ones?
Answer: 5C2=10. Choose positions for ones among five spots.
Flashcard 37: What is the formula for the number of ways to arrange n objects where some are alike?
Answer: p1!p2!...pk!n!. Divide by factorials of repeated object counts.
Flashcard 38: How many different 4-digit numbers can be formed using the digits 1-9?
Answer: 9×9×8×7=4,536. First digit excludes 0, others exclude previous digits.
Flashcard 39: What is the number of distinct permutations of the letters in LEVEL?
Answer: 2!2!5!=30. Two L's and two E's create identical arrangements.
Flashcard 40: Identify the number of ways to choose 0 objects from a set of 7.
Answer:
- By definition, C(n,0)=1 for any non-negative integer n.
Flashcard 41: What is the number of permutations where 3 items are identical and 2 are distinct?
Answer: 3!5!. Divides by 3! to account for identical items in permutations.
Flashcard 42: What is the number of 3-letter strings from 26 letters if repetition is allowed?
Answer: 263. Each position chosen independently with replacement.
Flashcard 43: What is 5C0?
Answer: 1. One way to choose nothing from any set.
Flashcard 44: What is the formula for combinations of r objects from n objects?
Answer: C(n,r)=r!(n−r)!n!. Divides by r! to eliminate order since combinations ignore arrangement.
Flashcard 45: What is the formula for circular permutations of n distinct objects?
Answer: (n−1)!. Fix one object to break circular symmetry.
Flashcard 46: What is the number of 3-letter strings from 26 letters if repetition is allowed?
Answer: 263. Each position chosen independently with replacement.
Flashcard 47: What is the Fundamental Counting Principle for m choices followed by n choices?
Answer: m⋅n total outcomes. Multiply choices at each step for sequential decisions.
Flashcard 48: What is the number of permutations of n distinct objects taken all at once?
Answer: n!. Each position has decreasing choices available.
Flashcard 49: What is the value of 1!?
Answer: 1!=1. Only one way to arrange a single object.
Flashcard 50: What is the Fundamental Counting Principle for m choices followed by n choices?
Answer: m⋅n total outcomes. Multiply choices at each step for sequential decisions.
Flashcard 51: What is the number of ways to assign president and vice president from 6 people?
Answer: 6P2=30. Officer positions require specific ordering.
Flashcard 52: Identify the rule: when choices are mutually exclusive, how do you count total outcomes?
Answer: Add the counts: m+n+⋯. Addition principle for disjoint events.
Flashcard 53: What is the number of outcomes when rolling a die and flipping a coin once?
Answer: 6⋅2=12. Independent events multiply their outcome counts.
Flashcard 54: How many ways can 2 identical balls be placed in 3 different boxes?
Answer:
- Stars and bars: C(2+3−1,3−1)=C(4,2)=6 placements.
Flashcard 55: How many ways can 4 different books be arranged on a shelf if 2 are always together?
Answer:
- Treat the pair as one unit: 3!×2!=12 arrangements.
Flashcard 56: What is the number of ways to assign president and vice president from 6 people?
Answer: 6P2=30. Officer positions require specific ordering.
Flashcard 57: What is 5P5?
Answer: 120. All arrangements of 5 distinct objects: 5!=120
Flashcard 58: What is the number of ways to arrange 8 objects in a circle?
Answer:
- Circular arrangements use (n−1)!=7!=5040 for 8 objects.
Flashcard 59: How many distinct permutations can be made from the letters in 'LEVEL'?
Answer:
- Calculate 2!2!5!=4120=30 with repeated letters.
Flashcard 60: How many ways can 8 people sit in a circle?
Answer:
- Fix one position to eliminate rotational symmetry.
Flashcard 61: How many ways can 2 identical balls be placed in 3 different boxes?
Answer:
- Stars and bars: C(2+3−1,3−1)=C(4,2)=6 placements.
Flashcard 62: How many ways can the letters in 'ACT' be arranged?
Answer:
- Three distinct letters give 3!=6 different arrangements.
Flashcard 63: What is the number of 3-letter strings from 26 letters if repetition is not allowed?
Answer: 26P3. Permutation formula for distinct selections.
Flashcard 64: What is the two-set inclusion–exclusion formula for ∣A∪B∣?
Answer: ∣A∪B∣=∣A∣+∣B∣−∣A∩B∣. Add sets, subtract overlap to avoid double-counting.
Flashcard 65: What is 5P5?
Answer: 120. All arrangements of 5 distinct objects: 5!=120
Flashcard 66: What does n! represent in counting methods?
Answer: The product of all positive integers up to n. Calculated as n×(n−1)×...×2×1.
Flashcard 67: How many ways can a president and vice president be selected from 5 candidates?
Answer:
- Order matters for president and vice president roles.
Flashcard 68: What is a combination?
Answer: A selection of items without regard to order. Order is not important in selections.
Flashcard 69: How many ways can 3 identical balls be placed in 4 different boxes?
Answer:
- Stars and bars: C(3+4−1,4−1)=C(6,3)=20 placements.
Flashcard 70: How many ways can a president and a vice-president be selected from 5 people?
Answer:
- Use permutations: P(5,2)=3!5!=20 ordered selections.
Flashcard 71: What is the number of distinct permutations of the letters in MISS?
Answer: 2!4!=12. Two S's create identical arrangements.
Flashcard 72: How many ways can you arrange the letters in the word 'BOOK'?
Answer: 2!4!=12. Two O's are identical, so divide by 2!.
Flashcard 73: What is the formula for permutations of n distinct objects taken r at a time?
Answer: nPr=(n−r)!n!. Accounts for unused objects in denominator.
Flashcard 74: How many distinct permutations can be made from the letters in 'LEVEL'?
Answer:
- Calculate 2!2!5!=4120=30 with repeated letters.
Flashcard 75: In a group of 10 people, how many ways can you choose a committee of 3?
Answer: (310)=120. Order doesn't matter when selecting committee members.
Flashcard 76: What is the formula for calculating combinations of n items taken r at a time?
Answer: C(n,r)=r!(n−r)!n!. Order doesn't matter, so divide by r! arrangements.
Flashcard 77: State the number of ways to arrange 3 books on a shelf.
Answer:
- Three distinct books can be arranged in 3!=6 ways.
Flashcard 78: What is the principle of multiplication in counting?
Answer: If one event has m outcomes and another has n, total is m×n. Sequential events multiply their outcome counts.
Flashcard 79: How many ways can 3 objects be selected from 5 distinct objects?
Answer:
- Using C(5,3)=3!2!5!=6⋅2120=10.
Flashcard 80: What is the formula for the number of subsets of a set with n elements?
Answer: 2n. Each element can either be included or excluded.
Flashcard 81: If a lock has 3 dials with 10 digits each, how many combinations are possible?
Answer: 103=1,000. Each dial has 10 choices, multiply together.
Flashcard 82: What is the number of circular arrangements of 7 distinct people around a round table?
Answer: (7−1)!=720. Fix one person's position to eliminate rotations.
Flashcard 83: How many ways can you arrange the letters in the word 'BOOK'?
Answer: 2!4!=12. Two O's are identical, so divide by 2!.
Flashcard 84: In a group of 10 people, how many ways can you choose a committee of 3?
Answer: (310)=120. Order doesn't matter when selecting committee members.
Flashcard 85: What is the number of ways to choose r items from n distinct items when order matters?
Answer: nPr. Selection where arrangement order matters.
Flashcard 86: What is the formula for calculating permutations of n items taken r at a time?
Answer: P(n,r)=(n−r)!n!. Order matters, so divide by unused positions (n−r)!.
Flashcard 87: What is the factorial definition of n! for a positive integer n?
Answer: n!=n(n−1)(n−2)⋯2⋅1. Product of all positive integers from n down to 1.
Flashcard 88: What is the binomial coefficient notation for combinations?
Answer: (rn). 'n choose r' notation for combination calculations.
Flashcard 89: What is the formula for the number of subsets of a set with n elements?
Answer: 2n. Each element can either be included or excluded.
Flashcard 90: What is the complement rule for counting: how do you count outcomes in set A using Ac?
Answer: ∣A∣=∣U∣−∣Ac∣. Count complement, subtract from universal set.
Flashcard 91: How many ways can you form a group of 4 from 10 people?
Answer: (410)=210. Same as choosing 4 from 10 people.
Flashcard 92: What is the number of ways to arrange n distinct objects in a circle (rotations considered identical)?
Answer: (n−1)!. Fix one position to eliminate rotational symmetry.
Flashcard 93: What is 5C5?
Answer: 1. One way to choose everything from any set.
Flashcard 94: If a lock has 3 dials with 10 digits each, how many combinations are possible?
Answer: 103=1,000. Each dial has 10 choices, multiply together.
Flashcard 95: What is the factorial of 5?
Answer:
- 5!=5×4×3×2×1.
Flashcard 96: Identify the number of ways to arrange 6 people if 2 must sit together.
Answer:
- Treat the pair as one unit: 5!×2!=240 arrangements.
Flashcard 97: What is 8C2?
Answer: 28. 2!6!8!=28⋅7=28
Flashcard 98: What is the value of 5!?
Answer:
- Compute 5!=5×4×3×2×1=120.
Flashcard 99: What is 7P3?
Answer: 210. 4!7!=245040=210
Flashcard 100: How many different outcomes are possible when tossing 3 coins?
Answer: 23=8. Each coin has 2 outcomes, multiply together.