A club of 10 members needs to elect a president, a vice-president, and a treasurer (three distinct roles). How many different leadership arrangements are possible?
AC(10, 3) = 120 — choose 3 members from 10; order doesn't matter since they're all officers
BP(10, 3) = 720 — choose and arrange 3 members in distinct roles; order matters
C10 × 3 = 30 — 10 candidates for each of 3 positions
D10! = 3,628,800 — consider all possible orderings of the full membership
When roles are distinct (president ≠ vice-president ≠ treasurer), the order of assignment matters. Alice as president / Bob as VP is different from Bob as president / Alice as VP. This is a permutations problem: P(10,3) = 10 × 9 × 8 = 720. If the roles were interchangeable (e.g., 'choose 3 committee members'), you would use C(10,3) = 120. The order-matters vs. order-doesn't-matter distinction changes the answer by a factor of 3! = 6.
Question 2 Multiple Choice
A bag contains 4 blue and 2 red marbles. Three marbles are drawn at random. What is the most efficient strategy for computing the probability that at least one red marble is drawn?
ACompute P(exactly 1 red) + P(exactly 2 red) separately and add the results
BCompute P(no red marbles) = C(4,3)/C(6,3) and subtract from 1
CCompute P(all red) and subtract from 1, since 'all red' is the complement of 'at least one red'
DMultiply the probability of drawing a red first by the probability of drawing another red
Complementary counting: P(at least one red) = 1 − P(no red). P(no red) = C(4,3)/C(6,3) = 4/20 = 1/5, so P(at least one red) = 4/5. The direct approach (option A) requires computing two separate cases. Option C is wrong — the complement of 'at least one red' is 'no red,' not 'all red.' Whenever 'at least one' appears, the complement is almost always the efficient path.
Question 3 True / False
When using the equally-likely outcomes formula, the same framework for counting — either both ordered or both unordered — must be applied consistently to both the numerator (favorable outcomes) and denominator (total outcomes).
TTrue
FFalse
Answer: True
Consistency between numerator and denominator is essential. If you count total outcomes as ordered sequences (permutations), you must also count favorable outcomes as ordered sequences. If you count total outcomes as unordered sets (combinations), favorable outcomes must also be unordered. Mixing the two inflates or deflates the probability by a factor of k! and produces a wrong answer.
Question 4 True / False
The probability that a 5-card hand contains at least one ace is most simply computed by summing the probabilities of getting exactly 1, 2, 3, and 4 aces.
TTrue
FFalse
Answer: False
Complementary counting is far simpler: P(at least one ace) = 1 − P(no aces) = 1 − C(48,5)/C(52,5). This is one calculation. Computing P(exactly k aces) for k = 1, 2, 3, 4 requires four separate combinatorial expressions — each of the form C(4,k)·C(48,5−k)/C(52,5) — and then summing them. Summing four such terms is the hard way to solve an 'at least one' problem.
Question 5 Short Answer
Explain why complementary counting (1 − P(none)) is usually the most efficient approach for 'at least one' probability problems.
Think about your answer, then reveal below.
Model answer: Because 'at least one' encompasses many sub-cases (exactly 1, exactly 2, exactly 3, ...) that each require separate counting, while 'none at all' is a single case that is often straightforward to count. The complement rule P(at least one) = 1 − P(none) replaces a multi-case sum with a single calculation.
For example, 'at least one ace in a 5-card hand' requires four separate combinatorial calculations summed together. P(no aces) = C(48,5)/C(52,5) is one calculation. The complement of 'at least one' is always 'zero of that thing,' which typically has only one case. This pattern holds broadly: whenever the direct calculation has many cases but the complementary event has just one, the complement is the efficient path.