A combination lock requires entering 3 digits from {1, 2, 3, 4, 5} in a specific order, with no repetition. How many valid codes are there, and which counting tool applies?
A10, using combinations C(5,3) — because you're just choosing 3 digits from 5
B60, using permutations P(5,3) — because a different sequence of the same digits opens a different lock
C6, using 3! — because only the arrangements of 3 chosen digits matter
D125, using 5³ — because each of 3 positions can independently be any of 5 digits
Order matters: the code 1-2-3 and 3-2-1 are different sequences that open different locks (or no lock). When order matters among distinct items selected without repetition, permutations apply: P(5,3) = 5×4×3 = 60. Combinations C(5,3) = 10 would count only the 10 distinct sets of digits, ignoring order — appropriate if the lock accepted any ordering of 3 chosen digits, which it does not. Option D ignores the no-repetition constraint.
Question 2 Multiple Choice
A committee of 4 students is chosen at random from a class of 25. What is the correct count of possible committees, and why?
AP(25,4) = 303,600 — because who holds which committee role matters
BC(25,4) = 12,650 — because the committee has no ordered roles; any group of 4 people is one committee regardless of selection order
C25⁴ = 390,625 — because each of 4 positions can be filled by any of 25 students
D25! — because all arrangements of the class must be considered
A committee is an unordered group — {Alice, Bob, Carol, Dan} is the same committee regardless of who was chosen first. When order does not matter, combinations apply: C(25,4) = 25!/(4! × 21!) = 12,650. Permutations P(25,4) would be correct if the 4 positions were distinguishable (president, vice-president, etc.), but a generic committee has no such ordering. This is the most common error in probability with combinatorics: using permutations when the selection is inherently unordered.
Question 3 True / False
The probability formula P(A) = (number of favorable outcomes)/(total number of outcomes) is valid for any sample space, regardless of whether the outcomes are equally likely.
TTrue
FFalse
Answer: False
This formula is valid only when all outcomes in the sample space are equally likely. For a fair die, each face has probability 1/6, so counting faces works. But a weighted die, or a sample space where outcomes have different probabilities (e.g., drawing from a bag with different numbers of colored balls and counting by color), requires assigning individual probabilities rather than counting uniformly. Applying the counting formula to a non-uniform space produces wrong answers. Recognizing when the equally-likely assumption holds is as important as knowing how to count.
Question 4 True / False
Choosing a 5-card poker hand from a deck of 52 and arranging 5 people in a row for a photograph both involve selecting 5 items from a larger set, so both problems use the same counting formula.
TTrue
FFalse
Answer: False
The distinction is whether order matters. For a poker hand, {A♠, K♠, Q♠, J♠, 10♠} dealt in any order is the same hand — combinations apply: C(52,5) = 2,598,960. For a photograph, the arrangement matters: Alice-Bob-Carol-Dan-Eve in a row is different from Eve-Alice-Bob-Carol-Dan. This requires permutations: P(52,5) or equivalently C(52,5) × 5! when selecting and then arranging. The surface similarity ('choose 5 from a set') conceals whether the order of selection or arrangement is meaningful.
Question 5 Short Answer
You want to find the probability of being dealt a 5-card hand with exactly two aces from a standard 52-card deck. Describe the counting approach: what goes in the numerator, what goes in the denominator, and why combinations are used rather than permutations.
Think about your answer, then reveal below.
Model answer: Denominator: C(52,5) = 2,598,960 — the total number of possible 5-card hands, where order doesn't matter. Numerator: C(4,2) × C(48,3) — choose exactly 2 of the 4 aces (C(4,2) = 6 ways), then choose 3 non-ace cards from the remaining 48 (C(48,3) = 17,296 ways). Combinations are used because a hand is an unordered set: the same 5 cards constitute the same hand regardless of the order they were dealt. P(exactly two aces) = (6 × 17,296) / 2,598,960 ≈ 0.0399.
The structure of the problem is: count favorable outcomes (hands with exactly 2 aces) divided by total outcomes (all 5-card hands). Breaking the numerator into two independent choices — which 2 aces, and which 3 non-aces — and multiplying uses the multiplication principle. Combinations appear throughout because card hands are inherently unordered; permutations would overcount by treating each ordering of the same 5 cards as a distinct hand.