Questions: Combinations and Unordered Selections

Question 1

A school needs to select 4 students to form a study group (no assigned roles). A second school needs to select a president, vice-president, secretary, and treasurer from 4 students. Both schools are choosing from 20 students. Which school's selection process has more possible outcomes?

• A. The study group — unordered selections always outnumber ordered arrangements
• B. The officer election — because the four chosen students can be arranged in 4! = 24 different role assignments, multiplying the count
• C. They are equal — both select exactly 4 students from 20
• D. The study group — because C(20,4) > P(20,4)

Answer: (B)

The officer election uses permutations P(20,4) = 20×19×18×17 = 116,280 because each ordering of the same 4 students represents a different outcome (different people in each role). The study group uses combinations C(20,4) = P(20,4)/4! = 116,280/24 = 4,845 because the same 4 students form one group regardless of order. Permutations are always ≥ combinations for r ≥ 2, so the ordered process has far more outcomes. Option D has the inequality backwards.

Question 2

How many ways can a committee of 3 be chosen from a group of 8 people?

• A. 8 × 7 × 6 = 336, because the order in which members are chosen matters
• B. C(8,3) = 8! / (3! × 5!) = 56
• C. 3! = 6, because there are 3 positions to fill
• D. P(8,3) = 336, divided by 2 = 168, because only some orderings matter

Answer: (B)

C(8,3) = 8!/(3! × 5!) = (8×7×6)/(3×2×1) = 336/6 = 56. A committee has no ordered roles, so {Alice, Bob, Carol} is the same committee as {Bob, Carol, Alice} — the 6 orderings of any trio are all the same outcome. Dividing P(8,3) = 336 by r! = 3! = 6 removes all this overcounting, giving 56 distinct committees. Option A uses the permutation formula, which overcounts by 6 for every committee.

Question 3

C(n, r) = C(n, n−r), which means choosing 3 items from 10 gives the same count as choosing 7 items from 10.

• T. True
• F. False

Answer: True

This symmetry holds because choosing r items to include is identical to choosing n−r items to exclude — the same partition of the full set is described from two perspectives. C(10,3) = C(10,7) = 120. This property is a useful computational shortcut: when r > n/2, compute C(n, n−r) instead, which involves smaller factorials. It also appears as the left-right symmetry in Pascal's triangle.

Question 4

The number of ways to arrange 5 books on a shelf equals the number of ways to choose 5 books from a collection of 5.

• T. True
• F. False

Answer: False

Arranging 5 books on a shelf is a permutation problem where order matters: P(5,5) = 5! = 120. Choosing 5 books from a collection of 5 — if all 5 must be chosen — gives C(5,5) = 1, because there is only one way to include all items. The arrangement question asks 'how many orderings of these items exist?', while the selection question (choosing all 5 from 5) asks 'in how many ways can we pick all of them?' — trivially one. These are completely different calculations.

Question 5

Why does C(n, r) = P(n, r) / r! ? What overcounting does dividing by r! correct for?

Model answer: P(n, r) counts ordered selections — it treats {Alice, Bob, Carol} and {Bob, Alice, Carol} and all other orderings as distinct outcomes. For every group of r specific items, P(n, r) counts all r! different orderings of that group. When we only care which items are chosen (not their order), each distinct group gets counted r! times by P(n, r). Dividing by r! removes this overcounting, leaving exactly one count per unique group. For r = 3: every combination is counted 3! = 6 times by P(n, r), so C(n, r) = P(n, r)/6.

The key insight is that P(n, r) overcounts by exactly r! for each combination — not approximately, but exactly. This is because every set of r items has exactly r! orderings, each counted once by the permutation formula. Combinations cancel this systematic overcounting uniformly, which is why the formula is exact and not an approximation.