A committee of 3 is chosen from 8 candidates. Which expression gives the number of possible committees?
A8 × 7 × 6
B8! / (3! × 5!)
C8! / 5!
D3! / 8!
C(8,3) = 8!/(3! × 5!) = 56. Option A (8 × 7 × 6 = 336) is P(8,3), which counts ordered selections — it overcounts because committee {Alice, Bob, Carol} and {Carol, Bob, Alice} are treated as different. Dividing by 3! = 6 corrects for this: 336/6 = 56. Option C omits the r! in the denominator; Option D inverts the formula entirely.
Question 2 True / False
C(10, 3) and P(10, 3) give the same count when choosing 3 people from 10.
TTrue
FFalse
Answer: False
P(10,3) = 720 counts ordered arrangements; C(10,3) = 120 counts unordered groups. C(10,3) = P(10,3)/3! = 720/6 = 120. They are equal only in the trivial case r = 1, where there is only one way to arrange a single item.
Question 3 Short Answer
Explain why C(10, 7) = C(10, 3) without computing either value.
Think about your answer, then reveal below.
Model answer: Choosing 7 items from 10 is equivalent to choosing which 3 items to leave out. Every selection of 7 items corresponds to exactly one set of 3 rejected items, so the two counts must be equal. This is the symmetry C(n,r) = C(n, n-r).
This symmetry — C(n,r) = C(n, n-r) — follows from the formula: n!/(r!(n-r)!) = n!/((n-r)!r!). The two denominators are the same, just written in reverse order. Conceptually, selecting a subset is the same decision as selecting its complement.