A teacher wants to select 4 students from a class of 20 to represent the school at a conference. How many ways can this selection be made, and which counting method applies?
AP(20, 4) = 116,280 — because we are choosing 4 students in sequence
BC(20, 4) = 4,845 — because the group of 4 is what matters, not who was chosen first
C20⁴ = 160,000 — because each of the 4 spots can be filled by any of 20 students
DC(20, 4) × 4! = 116,280 — because we must arrange the representatives alphabetically
The key question is: does order matter? A committee or representative group is an unordered set — {Alice, Bob, Carol, Dan} and {Dan, Carol, Bob, Alice} are the same group. So combinations apply: C(20, 4) = 20!/(4! × 16!) = 4,845. Option A is the permutation count, which would be correct if we were assigning roles (president, vice president, etc.) to the 4 students. Option D makes the mistake of combining both methods — if you use C(20, 4), you have already divided out all orderings among the chosen 4.
Question 2 Multiple Choice
A student claims: 'C(12, 5) must be different from C(12, 7) because you are choosing a different number of items.' Are they correct?
AYes — C(12, 5) = 792 and C(12, 7) = 792 happen to be equal only by coincidence
BNo — C(n, r) = C(n, n−r) always, because choosing 5 items is equivalent to choosing which 7 to leave out
CYes — the formula gives different values whenever r ≠ n−r
DNo — all combinations with the same n are equal regardless of r
The student is wrong. C(n, r) = C(n, n−r) is a fundamental identity, not a coincidence. Choosing 5 items from 12 is the same as choosing which 7 items to exclude — every selection of 5 uniquely determines a rejection of 7, and vice versa. So C(12, 5) = C(12, 7) = 792 exactly. This symmetry also provides a computational shortcut: when r > n/2, compute C(n, n−r) instead, which involves smaller factorials.
Question 3 True / False
The identity C(n, r) = C(n, n−r) holds for all valid values of n and r.
TTrue
FFalse
Answer: True
This identity follows directly from the formula: C(n, r) = n!/(r!(n−r)!) and C(n, n−r) = n!/((n−r)!r!). The denominators are the same product in different order, so the values are always equal. The combinatorial interpretation is equally clean: choosing r items to include is the same operation as choosing n−r items to exclude, since specifying one completely determines the other.
Question 4 True / False
In a race with 8 runners, the number of ways to determine the top 3 finishers (first, second, third place) is C(8, 3) = 56.
TTrue
FFalse
Answer: False
Order matters in a race — first, second, and third are distinct positions. {Alice 1st, Bob 2nd, Carol 3rd} is a different outcome than {Bob 1st, Alice 2nd, Carol 3rd}. So permutations apply: P(8, 3) = 8 × 7 × 6 = 336, not C(8, 3) = 56. The combination count 56 would be correct if we just wanted to know which 3 runners made the podium, without caring about their order — for example, selecting 3 runners to receive a generic medal. This is the essential judgment call: ordered → permutations, unordered → combinations.
Question 5 Short Answer
Explain why C(n, r) = P(n, r) / r!. What does dividing by r! correct for, and how does this derivation show that combinations count unordered selections?
Think about your answer, then reveal below.
Model answer: P(n, r) counts all ordered arrangements of r items chosen from n. But if we only care about which r items are chosen — not the order — then every distinct unordered set of r items has been counted r! times in P(n, r), once for each way to arrange those same r items. Dividing by r! cancels this overcounting, leaving exactly the number of distinct unordered selections. This is why C(n, r) = P(n, r) / r!: permutations overcount combinations by a factor of r!, the number of orderings of the selected items.
The derivation makes the formula meaningful rather than arbitrary. The key insight is that overcounting is systematic — every unordered selection is overcounted by exactly the same factor r! — so dividing by r! corrects all cases simultaneously. This also explains why combinations are always smaller than or equal to permutations: combinations disregard ordering information that permutations treat as significant.