A club of 10 members needs to elect a president, a vice-president, and a treasurer. How many ways can these three distinct offices be filled?
AC(10, 3) = 120, because we are choosing 3 people from 10
B10 × 10 × 10 = 1000, because each office independently has 10 candidates
CP(10, 3) = 720, because the offices are distinct and order of assignment matters
D3! = 6, because there are 3 offices to arrange
The offices are distinguishable (president ≠ vice-president ≠ treasurer), so assigning Alice as president and Bob as VP is different from assigning Bob as president and Alice as VP. This means order matters — it's a permutation problem. P(10, 3) = 10 × 9 × 8 = 720. Option A (combinations) would be correct if the three selected people formed an unranked committee with no distinct roles.
Question 2 Multiple Choice
How many distinct 3-letter arrangements can be formed using letters from {A, B, C, D, E} with no repetition?
A10, because C(5, 3) = 10
B60, because P(5, 3) = 5!/2! = 60
C125, because each of 3 positions has 5 choices
D15, because 5 × 3 = 15
We are choosing 3 letters from 5 and arranging them in order — order matters (ABC ≠ BAC), and no letter repeats. P(5, 3) = 5!/(5−3)! = 5!/2! = 5 × 4 × 3 = 60. Option A gives C(5,3) = 10, which counts unordered selections (combinations) — correct if you only cared which letters were chosen, not their order. Option C (125) counts arrangements with repetition, which is n^r = 5³, not applicable here.
Question 3 True / False
Arranging all n distinct objects in a complete sequence is a special case of permutations: P(n, n) = n!.
TTrue
FFalse
Answer: True
P(n, n) = n!/(n−n)! = n!/0! = n!/1 = n!. When you arrange all n objects (r = n), you're filling every slot, so the count is n × (n−1) × ⋯ × 1 = n!. This is exactly the 'number of ways to order n distinct objects' that factorial was introduced to count. Permutations generalize this: P(n, r) counts ordered arrangements when you only fill r of n slots.
Question 4 True / False
Permutations count arrangements where the order of selection does not affect the outcome.
TTrue
FFalse
Answer: False
This describes combinations, not permutations. Permutations count ordered arrangements, where swapping two elements produces a *different* outcome. For example, (Alice, Bob) and (Bob, Alice) are two distinct permutations of the same two people — because in a permutation context (like ranking first and second place), the order is what matters. Combinations count unordered selections, where {Alice, Bob} and {Bob, Alice} are the same.
Question 5 Short Answer
Explain why the formula P(n, r) = n!/(n−r)! has (n−r)! in the denominator.
Think about your answer, then reveal below.
Model answer: The numerator n! counts all ways to arrange all n objects in a complete sequence. But we only want to fill r slots — we use n, n−1, …, n−r+1 for those r choices, and the remaining (n−r) objects are not placed at all. Those unplaced objects can be in any order among themselves without changing the outcome, so we divide by (n−r)! to cancel the arrangements of the unused objects. The result n!/(n−r)! = n × (n−1) × ⋯ × (n−r+1) counts only the ordered choices for the r occupied slots.
The denominator isn't arbitrary — it precisely removes the overcounting caused by treating the unchosen objects as though they have meaningful positions. When r = n, (n−r)! = 0! = 1, so nothing is canceled and P(n,n) = n! as expected.